A cryptogram (part 4A)

Number Theory Level 4

A B C D + E F G H I J \begin{array} { l l l l l } & & A & B \\ & & C & D \\+ & E & F & G \\ \hline & H & I & J \\ \hline \\ \end{array}

In the above equation, each of the letters A , B , C , D , E , F , G , H , I A, B, C, D, E, F, G, H, I and J J represents a different one of the digits 0, 1, 2, ..., 9.

Note that A , C , E A, C, E and H H cannot be 0.

Suppose H = 9 H=9 . How many possible answers for H I J ‾ \overline{HIJ} ?

2 5 0 3 4 6 1

Chan Lye Lee by Chan Lye Lee , 44, Singapore

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1 solution

Alex Burgess
May 20, 2019

We have 2 ∗ H I J ≡ 0 ( m o d 9 ) 2 * HIJ \equiv 0 (mod 9) , so, H I J ≡ 0 ( m o d 9 ) HIJ \equiv 0 (mod 9) .

N u m b e r 981 = 7 ? ? + A B + C D , b u t 6 + 5 + 4 < 28 972 = 8 ? ? + A B + C D , b u t 6 + 5 + 4 < 17 963 = 40 + 51 + 872 954 = 20 + 61 + 873 945 = 10 + 62 + 873 936 = 10 + 52 + 874 927 = 10 + 53 + 864 918 = 7 ? ? + A B + C D , b u t 6 + 5 + 4 < 21 \begin{matrix} Number \\ 981 & = 7?? + AB + CD, but 6+5+4 < 28 \\ 972 & =8?? + AB + CD, but 6+5+4 < 17 \\ 963 & = 40 + 51 + 872 \\ 954 & = 20 + 61 + 873 \\ 945 & = 10 + 62 + 873 \\ 936 & = 10 + 52 + 874 \\ 927 & = 10 + 53 + 864 \\ 918 & = 7?? + AB + CD, but 6+5+4 < 21 \\ \end{matrix}

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