Find the Smallest of A and B: $A-B=98$ with Multiple of 19 Digit-sum

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The problem involves finding the smallest natural numbers A and B such that A - B = 98 and the digit sums of both A and B are multiples of 19. The discussion emphasizes the need to explore combinations of A and B that satisfy these conditions, leading to the conclusion that systematic testing of values is essential to identify the smallest valid pair. The solution requires a methodical approach to ensure both conditions are met simultaneously.

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$(1)A,B \in N,A-B=98$

(2)All the digit-sum of $A$ and $B$ are multiple of 19

please find the smallest of $A\,\, and \,\, B$
 
Last edited:
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Albert said:
$(1)A,B \in N,A-B=98$

(2)All the digit-sum of $A$ and $B$ are multiple of 19

please find the smallest of $A\,\, and \,\, B$

we have B mod 9 = x if sum of digits of a = 19 x
so (B + 98) mod 9 = x+ 8
for A and B both divisible by 19 we should have
so A Mod 9 is one less than B mod 9 with the proviso that b mod 9 = 0 => a mod 9 = 8

so smallest possible candidate should be digit sum of B= 38 and A = 19 as digit sum of B mod 9 is 1 more than digit sum of A mod 9
Smallest possible B = 29999 giving A = 30097 and it meets criteria
so B = 29999, A = 30097 is smallest solution,
 
Last edited:

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