Number Theory: Digitology
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1. In the correctly worked addition problem above, A, B, and C represent digits. What digit does C represent?

(A) 0
(B) 1
(C) 4
(D) 6
(E) 9

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No pair of two digit numbers adds up to something in the 200's. In other words, B has to be 1. There must have been a carry (if there wasn't a carry, the answer would be CC not BBC). When adding digits, the carry is always 1. So A+B+1 = 11. This means A+B = 10. So C = 0 and the answer is (A). BTW, A = 9.

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2. In the correctly worked addition problem above, A and B represent digits. What is A?

(A) 5
(B) 6
(C) 7
(D) 8
(E) 9

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Obviously there is a carry and A+1 = B. Now do trial and error: if B is 9 then A is 8 and 8+1 does equal 9. No other B will produce an A that is one less than B. The answer is (D).

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3. In the correctly worked addition problem above, A, B, C, D, E, and F represent six different digits. What is the largest possible value of C?

(A) 4
(B) 5
(C) 6
(D) 7
(E) 8

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There is no carry. If B, D, and E are 0, 1, and 2 (remember, they have to be different) then the biggest C you can have with no carry is 6. The answer is (C).

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4. In the correctly worked multiplication problem above, A, B, and C represent digits. The product of B and C is 32. What is A?

(A) 4
(B) 5
(C) 6
(D) 7
(E) 8

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Since B·C = 32 you know either B=4 and C=8 or B=8 and C=4. You carry a 3 so A·C=48. You can't have C=4 because then A would be 12 and 12 isn't a digit. So C must be 8 which makes A equal to 6. The answer is (C).

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5. A four-digit positive integer x is written as ABCD where A, B, C, and D are digits. Which equation correctly gives the value of A?

(A) A = (x – (B+C+D))/1000
(B) A = x/1000 – B/10 – C/100 – D/1000
(C) A = x – (100B + 10C +D)
(D) A = x + 100B + 10C + D
(E) A = x/1000 – B/100 – C/10 – D

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We know that x = 1000A + 100B + 10C + D (you learned that in elementary school but you didn't learn it with variables). Solving for A you get 1000A = x–100B–10C–D. Divide both sides by 1000 to get the answer (B).

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6. The sum of the digits of a four-digit number x is equal to N. The sum of the digits of 2x is also equal to N. The value of N could be:

(A) 16
(B) 18
(C) 20
(D) 22
(E) 24

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A little trial and error goes a long way on the SAT. If you double a 3 to a 6 the sum of the digits goes up by 3. But if you double a 6 to a 12, the sum of the digits goes down by 3. So if you double 3366, the sum of the digits is unchanged. Other examples are 4455 or 2277. In fact, if the sum of the digits is 9, 18, 27, or 36 then doubling the number will keep the sum the same. Anyway, the answer is (B).

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7. If x and y are numbers and A and B represent digits and 1/x = 0.0B and 1/y = 0.00A, what is Bx/(Ay)?

(A) 100
(B) 10
(C) 1
(D) 0.1
(E) 0.01

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Write 0.0B as B/100 and 0.00A as A/1000. This means that Bx=100 and Ay=1000. So Bx/Ay = 100/1000 = 0.1 or (D).

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8. If Q and R are integers and A and B represent digits and 1/Q = 0.0A and 1/R = 0.0B then one possible value of the product of A and B is:

(A) 6
(B) 12
(C) 15
(D) 18
(E) 20

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Using the technique from question 7, get Q = 100/A and R = 100/B. Since Q and R are integers, A and B must go evenly into 100 so they are 1, 2, 4, or 5. The product A·B could be a lot of things but it can't be 6, 12, 15, or 18. However, A·B could be 20. The answer is (E).

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9. In an alphabetic-to-numerical coding system, the single letters A-Z are given values of 0-25 respectively. In a three-letter word, the first letter has value µ, the second letter has value ß, and the third letter has value ∂. The value of a three-letter word is computed as follows:
word value = µ·262 + ß·261 + ∂·260
Example: CAT = 2 x 262 + 0 x 261 + 19 x 260 = 1371

What word has the value 783?

(A) ASK
(B) BED
(C) CUP
(D) DOG
(E) ELF

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Since 262 is 676, you have one of them in 783. This means the first letter is B. That's all you need. The answer is (B). To finish do: 1·676 + 4·26 + 3·1 = 783.

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10. Each of the three disks shown above rotates clockwise about its center. The disk on the right rotates at the rate of 90° per second. The disk in the middle rotates 90° each time the disk on the right completes one full rotation. The disk on the left rotates 90° each time the disk in the middle makes one full rotation. If the device starts with a reading of 0-0-0, what is its reading after 1 minute has elapsed?

(A) 2-3-0
(B) 2-3-1
(C) 2-3-2
(D) 3-2-2
(E) 3-3-0

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In 60 seconds the right-most disk makes 15 complete revolutions and is back on zero. The middle disk does 1 click for each of these 15 revolutions which means it does 3 complete rotations plus 3 clicks. So the middle disk lands on 3. The left-most disk does one click for each complete rotation of the middle disk so it lands on 3 also. The answer is (E).

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