5.3 Digits in Disguise

A cryptarithm is a puzzle where each letter stands for a single digit, distinct letters mean distinct digits, and a leading digit is never 0. We unpick them with number sense, parity and place value.

Worked Example

Solve \;AB\times 9 = CBA (a 2-digit number times 9 giving its reverse as a 3-digit number). Because the product is 3 digits, AB\times9\ge100, so AB\ge12. The units digit of AB\times9 must be A (the reversed number ends in A), and the leading digit C of the product is small. Testing the two-digit numbers whose nine-times reverses cleanly, only 91\times9 = 819 works: A=9,\,B=1,\,C=8.

Solve \;KN\times 3 = PKN (a 2-digit number times 3, with the product repeating KN behind a new leading digit). Tripling must leave the last two digits unchanged, so N has 3N ending in N, that forces N=0 or N=5. Checking, 50\times3 = 150 fits perfectly: K=5,\,N=0,\,P=1.

Try This Solve these multiplication cryptarithms:   (i) GH\times4 = HJ   (ii) RS\times5 = ST   (iii) AB\times6 = CCC   (iv) NM\times N = PPP   (v) GH\times H = 2K   (vi) PT\times6 = TTT.

    1. G=1, H=6, J=4   (16\times4 = 64).
    1. R=1, S=9, T=5   (19\times5 = 95).
    1. A=3, B=7, C=2   (37\times6 = 222).
    1. N=3, M=7, P=1   (37\times3 = 111).
    1. G=1, H=2, K=4   (12\times2 = 24).
    1. P=7, T=4   (74\times6 = 444).

Figure it Out: Reasoning & Cryptarithms

Practice

  1. If 2z34 is a multiple of 9, what is the digit z? Why are there two answers?
  2. Meera takes a number leaving remainder 8 on division by 12, and another that is 4 short of a multiple of 12. She claims their sum is always a multiple of 8. Examine the claim.
  3. When is the sum of two multiples of 3 a multiple of 6, and when not?
  4. Latha says any number divisible by 9 stays divisible by 9 when its digits are reversed. True? Are other digit shuffles also fine?
  5. If 5a632b is a multiple of 18, list all pairs (a,b).
  6. If 6p1q2 is divisible by 44, list all pairs (p,q).
  7. Find three consecutive numbers where the first is a multiple of 2, the second of 3, the third of 4. How often do they occur?
  8. Write five multiples of 36 between 45000 and 47000.
  9. The middle number of five consecutive even numbers is 5p. Express the other four in terms of p.
  10. Write a 6-digit number divisible by 15 such that reversing its digits gives a number divisible by 6.
  11. Arjun claims some multiples of 11 stay multiples of 11 when doubled, but others don’t. Is he right?
  12. Always / sometimes / never?   (i) The product of a multiple of 6 and a multiple of 3 is a multiple of 9.   (ii) The sum of three consecutive even numbers is divisible by 6.   (iii) If abcdef is a multiple of 6, then badcef is a multiple of 6.   (iv) 8(7b-3)-4(11b+1) is a multiple of 12.
  13. Choose any 3 numbers. When is their sum divisible by 3?
  14. Is the product of two consecutive integers always a multiple of 2? Three consecutive a multiple of 6? Four? Five?
  15. Solve:   (i) DE\times D = FFF   (ii) WOW\times5 = MEOW.
  16. Which Venn diagram captures the relationship between multiples of 4, 8 and 32?
  1. 2+z+3+4 = 9+z must be a multiple of 9, so z=0 or z=9, two answers, because both 9 and 18 are multiples of 9 reachable by a single digit.
  2. Let a = 12n+8 and b = 12m-4. Then a+b = 12(n+m)+4 = 12k+4, which is not always a multiple of 8 (e.g. k=1 gives 16, which is, but k=2 gives 28, which is not). Claim false.
  3. Two multiples of 3 are 3m, 3n; their sum 3(m+n) is a multiple of 6 only when m+n is even.
    1. Yes, reversing the digits leaves the digit sum unchanged, so divisibility by 9 survives. (ii) Any rearrangement of the digits keeps the same digit sum, so it stays a multiple of 9.
  5. Need divisibility by 2 (b even) and by 9 (5+a+6+3+2+b = 16+a+b a multiple of 9, so a+b=2 or a+b=11). Working through even b gives (a,b) = (2,0),\,(0,2),\,(7,4),\,(5,6),\,(3,8),\,(9,2).
  6. Need divisibility by 4 (last two digits q2 \in \{12,32,52,72,92\}, all \equiv even-and-by-4 cases) and by 11 (alternating sum of 6p1q2 is 2-q+1-p+6 = 9-(p+q), which must be 0 or \pm11, so p+q=9). The pairs are (p,q) = (0,9),\,(2,7),\,(4,5),\,(6,3),\,(8,1).
  7. One set is 2,3,4. The pattern repeats every \operatorname{LCM}(2,3,4)=12, so the next is 14,15,16, then 26,27,28, and so on, every 12 numbers.
  8. A multiple of 36 must be divisible by 4 and 9. Examples: 45036,\ 45072,\ 45108,\ 45144,\ 45180 (many others exist).
  9. 5p-4,\ 5p-2,\ 5p+2,\ 5p+4.
  10. Divisible by 15 means ending in 0 or 5; reversing must give a 6-digit number divisible by 6 (so even, leading digit even and non-zero). With last digit 5 and first digit even, e.g. 400035 works (it is divisible by 15, and its reverse 530004 is divisible by 6).
  11. False. If n = 11k then 2n = 22k = 11(2k), doubling a multiple of 11 always gives a multiple of 11.
    1. Always (a multiple of 6 carries a 3, a multiple of 3 carries another, so the product holds 3\times3). (ii) Always (2k+(2k+2)+(2k+4) = 6k+6 = 6(k+1)). (iii) Always (swapping digit pairs of equal place-parity keeps both the units digit and the digit sum, so divisibility by 2 and 3, hence 6, survives). (iv) Never: 8(7b-3)-4(11b+1) = 56b-24-44b-4 = 12b-28 = 12(b-2)-4, always 4 short of a multiple of 12.
  13. With remainders R_1,R_2,R_3 on division by 3, the sum is divisible by 3 exactly when R_1+R_2+R_3 is 0,3 or 6, i.e. all three remainders equal, or they are 0,1,2 in some order.
  14. Two consecutive integers: one is even, so the product is a multiple of 2. Three consecutive: contain a multiple of 2 and a multiple of 3, so a multiple of 6. Four consecutive: a multiple of 24 (2\times3\times4). Five consecutive: a multiple of 120 (2\times3\times4\times5).
    1. D=3, E=7, F=1 (37\times3 = 111). (ii) W=5, O=7, M=2, E=8 (575\times5 = 2875).
  16. (iv), every multiple of 32 is a multiple of 8, and every multiple of 8 is a multiple of 4, so the sets nest: multiples of 32 \subset multiples of 8 \subset multiples of 4.