Finger Multiplication: A Mathematical Excursion Randall Baumback
“I know my ones. I know my twos. I know my threes... Ah, the finger man.” An unforgettable exchange between a defiant, remedial student and a confident, driven math teacher in the movie Stand and Deliver [1]. This back‐and‐forth presented a teachable moment for the determined teacher‐ giving way to a class presentation of the finger manipulation for multiplying by nine. A technique familiar to most every elementary school student. There are several finger multiplication techniques. Let’s start with the “multiplying by nine,” [2] as it is commonly known, generalize and expand from there, and see where it takes us. Multiplying by nine on our fingers is simple to implement and easy to understand. Using to represent a finger and as the dropped finger, we can represent, for example, 3 x 9 = 27 by positioning our hands in front of us with palms facing away and dropping the third finger from the left in the following way.
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9
10
The two circles to the left of represent the ten’s place digit and the seven circles to the right represent the one’s place digit. The representation yields the correct answer 27. Why does it always work? Let n be the number multiplied by nine. Dropping the nth finger results in (n ‐ 1) fingers to the left and (10 ‐ n) fingers to the right. The value of these fingers is 10(n ‐ 1) + (10 ‐ n) = 9n. Voila! Have you ever seen this finger manipulation done with a number other than nine? Why only multiplying by nine? What’s so special about nine? Simply, because we have ten fingers. This makes it natural to count and calculate in a base ten number system and dropping one of the fingers leaves nine fingers. There you have it. I guess you can argue we really only have eight fingers since the thumbs don’t count as fingers. In that case, it would be natural for us to count and calculate in base eight. It would seem to follow that we could use the same finger technique to multiply by seven. Let’s try it. How about 5 x 7?
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8
Let’s see. 5 x 7 = 43? Remember, base eight. The 4 is in the eight’s place and 3 Finger Multiplication page 1 of 13 A Mathematical Excursion
ones. So, 4 x 8 + 3 and we’re back to base ten. Okay, 438 = 3510. It works. Neat! What if we met an alien and shook it’s hand while noticing seven figures (on each of it’s four hands) or maybe it had three fingers on all eight hands. Does this finger manipulation trick work for them or any other combination of fingers and hands? Sure. Why not? Let’s generalize the base (number of fingers). Let b be any integer base greater than 1, and therefore, multiply (b ‐ 1) by any integer, n, such that 1 < n < (b‐ 1).
1
2
3
... ...
n
... ...
b-1
b
There are (n ‐ 1) fingers to the left and (b ‐ n) to the right. The product is ... how can I express it? Interpret the result as two numbers side‐by‐side base b to form the overall number base b, for example, (x)(y)b (and not the multiplication of two numbers). (n ‐1)(b ‐ n)b = [b(n ‐ 1) + (b ‐ n)]10 = [bn - n]10 = [n(b ‐ 1)]10. Exactly what we expected. So, there you have it. The “multiply by nine” technique works for any integer greater than 1. If this finger manipulation trick is so good, why don’t we see it used daily in math classes across the country? What’s the downside? In a word, impractical. For nine, it’s great. For smaller numbers, it’s not unreasonably tedious. For large numbers, not so good. So, it’s just too limited for practical purposes. For large b, we have the standard “better” algorithm we all learned in school. For example, what if the base is 130, and we want to compute 93 x 129. Either we pull out a calculator or the standard method below where there’s no need to do base conversion (since we work in base 10). 1 2 9 x 9 3 3 8 7 1 1 6 1 1 1 9 9 7 Nonetheless, the finger manipulation trick is interesting from a mathematics standpoint and is a fun way to learn base conversion. We’ll come back to the connection of the finger manipulation and base conversion later seeing it’s value with a few surprising links in elementary number theory. Of course, both the finger manipulation and standard algorithms for multiplication are rendered useless with limited knowledge of the basic multiplication facts. I guess it’s always good advice to study early and often the Finger Multiplication page 2 of 13 A Mathematical Excursion
multiplication table. Putting that aside for the moment, is there another finger manipulation for multiplying? You know the answer to that question. Oh, where is it? First paragraph. “There are several finger multiplication techniques.” Let’s take a look at another procedure for multiplying on your fingers [3]. Guaranteed, this one is a crowd pleaser both for children and adults, alike. Again, we’ll use circles to represent fingers. Shaded ones suggesting importance and touching. But this time position our hands in front of us with palms facing inward and thumbs on top. Now comes the leap of faith. Let’s suppose everyone over the age of 21 (or, maybe, 9) knows the multiplication facts for the ones through the fives. It’s the big numbers, the sixes through the tens that are problematic. This clever device takes care of the problem. Furthermore, it is easy to use and pleasing to understand. Since we’re only concerned about the fives through the tens, we will number our fingers in order from bottom to top (thumbs are 10) as shown below. 10 9 8 7 6
10 9 8 7 6
The best way to learn this finger multiplication device is to see a couple of examples and then try it for yourself. Let’s consider the product a x b. Remember, a and b are integers between 6 and 10, inclusive. It’s a four step algorithm. First, touch the a finger on the left hand with the b finger on right hand. Second, count the fingers that are touching and below (let’s call these the bottom fingers) giving each a value of ten. Hold on to that thought. Third, multiply the count of the remaining fingers above on the left hand by the remaining fingers above on the right hand (and let’s call these the top numbers) . Don’t worry, you’ll get it. Fourth, combine the result of steps two and three and you’ve got your answer. Let’s give it a go. What is 7 x 8? First, touching our seven and eight fingers. 10 9 10 8 9 7 8 6 7 6 Second, five fingers touching and below for a total bottom value of 50. Third, left above three fingers x right above two fingers giving 3 x 2 = 6. Fourth, combing the Finger Multiplication page 3 of 13 A Mathematical Excursion
results in steps two and three produces the product 7 x 8 = 50 + 6 = 56. Not bad. What do you think is the most troublesome product in this system? Nope, not the largest. Instead, the smallest product, 6 x 6. Why?, you ask. You’ll see it requires knowing the largest multiplication fact (since more fingers left on top) and, although still simple, requires the most challenging sum. 10 9 8 7 6
10 9 8 7 6
Two bottom fingers. That’s 20. Now, top fingers, 4 x 4 = 16. So, 6 x 6 =20 + 16 = 36. So, why does it work? Suppose the numbers we’re multiplying are a and b. That makes the value of the bottom fingers 10[(a ‐ 5) + (b ‐ 5)] = 10(a + b) ‐ 100. Multiplying top fingers on each hand together yields (10 ‐ a)(10 ‐ b) = 100 ‐ 10(a + b) + ab. Adding the results, produces [10(a + b) ‐ 100] + [100 ‐ 10(a + b) + ab] = ab, the desired product. Maybe, we can generalize this device beyond sixes through tens. Do you feel lucky? I’ll give you a hand. You’re going to need it‐ literally. Let’s first consider a partial generalization but it’s going to take more than two hands. Visualize multiple pairs of hands stacked one above the other. Starting with the bottom pair, number the fingers the same as in the previous scheme. Continue numbering the fingers in each successive pair of hands as consecutive integers as shown below.
Finger Multiplication page 4 of 13 A Mathematical Excursion
5(n+1)
5(n+1) nth pair
your hands
my hands 15 14 13 12 11 10 9 8 7 6
15 14 13 2nd pair other hands 12 11 10 9 8 1st pair 7 6
We need to impose the constraint that the numbers being multiplied must be on the same pair of hands. For instance, 12 x 15 or 21 x 25 or 66 x 68 are all okay. 9 x 13 or 23 x 39 are not okay. We’ll use a modification of the two‐hand device. However, we’ll need to apply it over and over as many times as needed. Think of it as an iterative loop process. While it sounds like a recursive algorithm better left to a computer, the calculations are simple and a pattern will emerge that allows for shortcuts. Initially, it’s a good idea to create a table to organize everything and to better see a pattern develop. The first column (Bottom) in our table will identify the count of the bottom fingers. The second column (Value) will show the total value of the bottom fingers. The third column (Top) will show the top numbers to be multiplied and then added to the bottom value. The iterative nature becomes clear from our third (Top)column observation. These numbers will be larger than the two‐hand device can handle since our original multiplicand will be substantially larger than ten. As a result, we’ll start the process anew with the two top numbers from the the third column. Eventually, we’ll need to combine all our results. First things first, though. In the two‐hand device, the bottom fingers were always worth 10 each. This is not true for the multi‐hand device. We need to establish the worth of the bottom fingers before we do all else. The good news is the worth is easy to figure. Looking back on the two‐hand device, the worth of each bottom finger was ten because there was a pair of “imaginary” or missing hands with ten fingers below the pair of hands numbered six through ten. Using the same reason, establishing the value of the bottom Finger Multiplication page 5 of 13 A Mathematical Excursion
fingers is a function of the number of pairs of hands (including the “imaginary’ pair) that are below the pair of hands with the touching fingers. Once this worth is established, the top numbers are found by subtracting each multiplicand from the worth just like we did in the two‐hand device (10 ‐ a and 10 ‐ b). Let’s start with 52 x 54 =2,808. 55 54 53 52 51
55 54 53 52 51
How many sets of hands are below your hands? There are 50/5 = 10 pairs. Therefore, the bottoms fingers are worth 10 x 10 = 100. What are the top numbers? 48 and 46 since 100 ‐ 52 = 48 and 100 ‐ 54 = 46. Creating the table should help us continue. Bottom
Value
Top
6
600
48 x 46
4
360
42 x 44
6
480
38 x 36
4
280
32 x 34
6
360
28 x 26
4
200
22 x 24
6
240
18 x 16
4
120
12 x 14
6
120
8 x 6
4
40
2 x 4
table 1 What patterns do you notice? In column Bottom we have an alternating sequence of two numbers that sum to 10. Once you have the first number the rest are predictable. What about column Value? Each entry is found by multiplying column Bottom by 100, 90, 80, ..., 10 as you work your way down the column Bottom. Do you see the doubly cyclic patterns in column Top? The ones digit in both numbers alternate vertically down the rows between two numbers that sum to 10 (hum, recurring trend). Additionally, the tens digit in both numbers (are the same) and Finger Multiplication page 6 of 13 A Mathematical Excursion
they repeat, drop by one, repeat, drop by one, etc. as they cascade down the rows until both numbers become single digits. Another way of expressing these patterns in column Top is to assign one set of alternating rows one color and the other set of alternating rows a different color as you move down the rows until both numbers become single digits. It becomes clear that both numbers of like color decrease by 10 (there it is again) from top to bottom rows. We’ve produced the table. Now what? Simply sum the numbers in column Value and add to that the last product found in the table. Don’t forget our assumption... everyone knows the multiplication facts for the ones through the fives. Did this multi‐hand device work? I like our chances since the sum of column Value must end in 0 and our last product is 8 and isn’t 52 x 54 = 2,808? Let’s not leave it to chance, though. 10(60 + 36 + 48 + 28 + 36 +20 + 24 + 12 + 12 + 4) + 2 x 4 = 10 x 280 + 8 = 2,808. Whew! Maybe we were lucky. Does the the multi‐hand scheme always work under our constraint? I hope so. Only one way to find out. Let’s generalize our multiplying numbers. For simplicity, let’s stick to two‐digit numbers and feel confident that similar reasoning but more messy algebra will work for numbers with three or more digits. Let X =ab and Y =ac be two two‐digit numbers (not two products). Suppose we wish to multiply X and Y, ie., XY = ab x ac. The ten’s digit is the same because of our constraint that the numbers must be on the same pair of hands. Unfortunately, there are two cases to consider. Case (i): b and c are both less than or equal to 5; and case (ii): both b and c are greater than 5. One digit above 5 and the other not is prohibited since the touching fingers are constrained to the same pair of hands. For case (i), each bottom finger has a value of the number of pairs of hands below times 10 (because ten fingers per pair). In the general case, the value is 10(10a/5) = 20a. For case (ii), the value of each finger is 10(10a/5 + 1) = 20a + 10. With so many other problems in the world to solve, let’s concentrate on case (i) and assume case (ii) can be approached in the same way. You know the mathematicians creed, “if it works n times, then it will work (n + 1) times.” The table worked before... For XY = ab x ac, the value of the fingers below is 20a.
Finger Multiplication page 7 of 13 A Mathematical Excursion
Bottom
Value
Top
b+c
20a(b + c)
(20a ‐ ab) x (20a ‐ ac)
10 - (b + c)
(20a - 10)(10 - (b + c))
:
b+c
(20a ‐ 20)(b + c)
:
10 - (b + c)
(20a ‐ 30)(10‐(b + c))
:
|:::n::
:
:
10 - (b + c)
10
bxc
table 2 The multi‐hand algorithm is the sum of column Value + (b x c). Hereafter, b x c = bc. Dig in . . . 10(b + c)[2a + (2a ‐ 2) + (2a ‐ 4) + ... + 0] + 10(10 ‐ (b + c)) [(2a ‐ 1) + (2a ‐ 3) + ... + 1] + (b x c) = 20(b + c)[a + (a ‐ 1) + (a ‐ 2) + ... + 0] + 10(10 ‐ (b + c))[(2a ‐ 1) + (2a ‐ 3) + ... + 1] + (b x c)
note 1: [a + (a ‐ 1) + (a ‐ 2) + ... + 0] is the sum of the consecutive integers from 1 to a which sum to a(a + 1)/2. note 2: [(2a ‐ 1) + (2a ‐ 3) + ... + 1] is the sum of the consecutive odd integers from 1 to (2a ‐ 1) which sum to ((2a)/2)2 = a2. = 20a(b + c)(a + 1)/2 + 10a2(10 ‐ (b + c)) + bc = 10a(a + 1)(b + c) + 10a2(10 ‐ (b + c)) + bc = 10a[(a + 1)(b + c) + a(10 ‐ (b + c)) + bc = 100a2 + 10ab + 10ac + bc = (10a + b)(10a + c) Yes! (accompanied with fist pump). We recognize the last expression as the base ten representation of the two‐digit number ab multiplied by the two‐digit number ac. Don’t get me wrong. There’s no claim the multi‐hand manipulation is better than the the old classic multiplication algorithm. I just wanted to know if the crowd pleasing two‐hand manipulation could work for larger numbers provided we had more hands. And now we know. It can. Taking advantage of the patterns revealed in the table allowing for shortcuts,and with some practice, the multi‐hand Finger Multiplication page 8 of 13 A Mathematical Excursion
manipulation is an interesting alternative to the standard multiplication algorithm. Okay, one more for the road. How about 33 x 35? The three and five fingers are touching and they are each worth 60 (30/5 x 10) for a total value of 480 (8 x 60). Next 320 (8 x 40). Hum, a differnce of 160 gives us 480 + 320 + 160 = 960 for the Bottom. Now the Top is 2 (10‐8). So, 100 (2 x 50) and then 60 (2 x 30) and last 20 (difference of 40) giving 180 (100 + 60 + 20). Total Value is 1140 (960 + 180). Add in the 15 (3 x 5). For a grand total of 1,155 (1140 + 15). Right? This finger multiplication has been some trip. From the familiar surroundings of “multiplying by nines” finger manipulation in base ten to it’s neighbor in any base. Then on to a new culture with the introduction of the “two‐hand” algorithm. Finally, ending up a long way from home, yet with the comfort of a relative , the “multi‐hand” scheme. Maybe, it’s time to return to our roots. Back to the good ol’ neighborhood where it all started‐ the relationship of calculation and base. An inherent, fundamental connection. There are a lot of numbers out there to explore at the same time. We’ll continue to limit our exploration to non‐negative integers for now. Even then, a lot of numbers. So let’s partition those into three smaller groups of evens, odds and primes. I know, technically, those sets aren’t any smaller than the non‐negative integers but can be easier to investigate separately. Think of it as painting two rectangle rooms with the same surface area but one room doesn’t require a ladder to paint. Let’s get started. By definition, all positive even integers can be generated by multiplying q by 2 for q = 1, 2, 3, . . . Take 14, for example. 14 = 2 x 7. Returning to the finger manipulation and working backward to produce 14, we’ll use base 8.
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From above we see 168 = 1 x 8 + 6 = 14. Observe, 6 + 8 = 14. Let’s take 56 this time. We’ll skip the diagram. 56 = 2 x 28. 1(27)29 = 1 x 29 + 27. Observe, 27 + 29 = 56. By the way, the notation (ab) represents a single place in this particular base. In this case, (27) is in the one’s place. Be patient, Rome wasn’t built in a day. How about 98 = 2 x 49. 1(48)50 = 1 x 50 + 48 = 98. Observe, 48 + 50 = 98. It appears every even positive integer can be witten as a sum of two consecutive even integers. Let’s try another. 12 = 2 x 6. 157 = 1 x 7 + 5 = 12. Notice, this time we have a sum of two consecutive odd integers. No problem. A minor adjustment. Also notice we always chose 2 as one factor. Finger Multiplication page 9 of 13 A Mathematical Excursion
Integer
Product
Base
Left
Right
Sum
14
2 x 7
8
1
6
6 + 8
56
2 x 28
29
1
27
27 + 29
98
2 x 49
50
1
48
48 + 50
12
2 x 6
7
1
5
5 + 7
:
:
:
:
:
:
2n
2 x n
n + 1
1
n ‐ 1
(n ‐ 1) + (n + 1)
table 3 Using the same mechanism on odd integers as we did with even integers, let’s derive a table (table 4) and see what we get. Let’s begin with 15 = 3 x 5. Applying base 6 leaves 2 on left and 3 on right. As before, 236 = 2 x 6 + 3 = 15. Here’s where experimentation comes into play. Decrease the value left (2) by one while increasing the base (6) by one. Repeat this until the value left becomes 1. Now, consider the series of consecutive odd integers from the value right to the base value for this last repetition (3 + 5 + 7). Observe, this sum is equal to the original odd integer. The difference between the odd sum from 1 to 7 and the odd sum from 1 to 1 is the same as the odd sum from 3 to 7 equally 15. Remembering the formula for the sum of the first n odd positive integers is n2 , we conclude that 15 = (number of odd terms from 1 to 7)2 ‐ (number of odd terms from 1 to 1)2 = [(7 + 1)/2]2 - [(1 + 1)/2]2 = 42 - 12. What would happen if we used 15 = 1 x 15? Applying base 16 leaves 0 on left and 15 on right. 0(15)16 = 0 x 16 + 15 = 15. This time we increase value left by one while decreasing the base by one. So, we use the odd series from 1 to 15. This sum doesn’t produce the original integer 15. However, we see that 82 ‐ 72 = 15. Let’s try a prime number, say, 19. Of course, by definition, we can only use 19 = 1 x 19. Using the same steps as we did if 15 yields (1 + . . . + 19) and 102 ‐ 92. How about one more non‐prime? Okay then, 49. Will a perfect square work? If we use 49 = 1 x 49, we produce (1 + . . . + 49) and 252 ‐ 242. Let’s check it to be sure. 625 ‐ 576 = 49. We knew it would. Now, the big test. 49 = 7 x 7. Use base 8. Left 6 and right 1. 618 = 6 x 8 + 1. So far, so good. Repeatedly reducing the 6 while simultaneously increasing the 8 and the last product pair of (left, base) values are (1, 13). Forming the odd sum from 1 to 13 (1 + 3 + 5 + 7 + 9 + 11 + 13) = 49. Nice. Since the series starts with 1, it sums like a 1 x n product. [(13 + 1)/2]2 ‐ [0/2]2 = 72 ‐ 02 . Mission accomplished but in a trivial form.
Finger Multiplication page 10 of 13 A Mathematical Excursion
Integer
Product
Base
Left
Right
Sum
Difference
15
3 x 5
6
2
3
3 + 5 + 7
42 ‐ 12
15
1 x 15
16
0
15
1 + ... + 15
82 ‐ 72
19
1 x 19
20
0
19
1 + ... + 19
102 ‐ 92
49
7 x 7
8
6
1
1 + ... + 13
72 ‐ 02
:
:
:
:
:
:
:
2n + 1
_
_
_
_
_
_
table 4 I won’t claim there are an endless number of observations we can make from tables 3 and 4 but there are many. Since the ideas were born from our fingers, a “handful” are listed below. Conjecture 1: Every even positive integer can be witten as a sum of two consecutive even or two consecutive odd integers. Conjecture 2: Every multiple of four positive integer can be witten as a difference of two consecutive even or two consecutive odd integers that are squared. Conjecture 3: Every non‐prime odd integer can be witten as a sum of consecutive integers. Conjecture 4: Every non‐perfect‐square odd integer can be witten as a difference of two consecutive integers that are squared. Conjecture 5: Every prime number greater than 2 can be written as a difference of two consecutive integers that are squared. Proof of Conjecture 1: Let z be an even positive integer such that z = 2n for n = 1, 2, 3, . . . Let z1 = (n ‐ 1) and z2 = (n + 1). Then z2 ‐ z1 = (n + 1) ‐ (n ‐ 1) = 2 implies z1 and z2 are two consecutive even integers when n is odd or consecutive odd integers when n is even. Therefore, z = 2n = (n ‐ 1) + (n + 1) = z1 + z2 is a sum of two consecutive even or two consecutive odd integers. Finger Multiplication page 11 of 13 A Mathematical Excursion
Proof of Conjecture 2: Let z be an even positive integer such that z = 4n for n = 1, 2, 3, . . . Let z1 = (n ‐ 1) and z2 = (n + 1). Then z2 ‐ z1 = (n + 1) ‐ (n ‐ 1) = 2 implies z1 and z2 are two consecutive even integers when n is odd or consecutive odd integers when n is even. Therefore, z = 4n = (n + 1)2 ‐ (n ‐ 1)2 = z22 ‐ z12 is a difference of two consecutive even or two consecutive odd integers that are squared. Proof of Conjecture 3: Let z be a non‐prime odd integer, z = 2n + 1 for n = 1, 2, 3, . . . that don’t produce a prime number. Let z1 = n and z2 = n + 1. Then z2 ‐ z1 = (n + 1) ‐ (n ) = 1 implies z1 and z2 are consecutive integers. Therefore, z = 2n + 1 = n + (n + 1) = z1 + z2 is a sum of two consecutive integers. Proof of Conjecture 4: Let z be a non‐perfect‐square odd integer, z = 2n + 1 for n = 1, 2, 3, . . . that don’t produce a perfect square. Let z1 = n and z2 = n + 1. Then z2 ‐ z1 = (n + 1) ‐ (n ) = 1 implies z1 and z2 are consecutive integers. Therefore, z = 2n + 1 = (n + 1)2 ‐ n2 = z22 z12 is a difference of two consecutive integers that are squared. Proof of Conjecture 5: Let p be a prime number greater than 2 implying odd p = 2n + 1 for some n = 1, 2, 3, . . . Let z1 = (p ‐ 1)/2 and z2 = (p + 1)/2. Then z2 ‐ z1 = (p + 1)/2 ‐ (p ‐ 1)/2 = 1 implies z1 and z2 are consecutive integers. Therefore, p = [(p + 1)/2]2 ‐ [(p ‐ 1)/2]2 = z2 2 ‐ z12 is a difference of two consecutive integers that are squared. The proof (of each conjecture) is in the pudding, so to speak. Once the base conversion pattern is observed and tied to sums or differences, the generating formulas make the proofs nearly self‐evident. But like most excursions, it’s the journey not the destination that matters. It’s the discovery of each conjecture through a seemingly unrelated connection to a relationship of finger multiplication and base conversion that is exciting. While this excursion of finger multiplication has come to an end, not to worry, there are many more explorations ahead. Finger Multiplication page 12 of 13 A Mathematical Excursion
References 1. 2.
3
4. 5. 6.
T. Keogh, ʺStand and deliver,” n.d., (accessed August 13, 2010). M. Eyler and V. Valero,” Multiplying by nine using your hands,” YouTube. November 13, 2008, (accessed Agusust 13, 2010) J. Dalziel, “Finger trick multiplication times tables 6‐10,”Lancaster University, 2005, (accessed August 13, 2010) G Flegg, Numbers: their history and meaning, Dover, 1983 H. Eves, History of mathematics, Holt, Rinehart and Winston, 1969 I. Niven and H. Zuckerman, An introduction to the theory of numbers, Wiley & Sons, 1972
Comment 1.
Stand and Deliver (1988), Warner Brothers, directed by Ramon Menendez , written by Ramon Memendez and Tom Musca: “Edward James Olmos’s Oscar‐nominated performance energizes this true‐ life story of a Los Angeles high school teacher who drives his students on to excellence at calculus. Based on a true story, this inspiring American Playhouse production stars Edward James Olmos as a high school teacher who motivated a class full of East L.A. barrio kids to care enough about mathematics to pass an Advanced Placement Calculus Test. Not exactly a variation of To Sir, With Love, the film concerns itself with assumptions and biases held by mainstream authorities about disadvantaged kids, and Olmos’s efforts to keep his students coolheaded enough to prove them wrong. Olmos, virtually unrecognizable as the pudgy, balding instructor, gives a career performance in this fine piece directed by Ramón Menéndez, and written by the director and Tom Musca.” –Tom Keogh
Randall Baumback James Logan H.S., Union City, CA USA Ohlone College, Fremont, CA USA
Finger Multiplication page 13 of 13 A Mathematical Excursion
