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

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To find 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, one must explore various combinations of A and B that satisfy these conditions. The digit sum requirement significantly narrows down the possibilities, as both numbers must yield sums divisible by 19. Through systematic testing of values, the smallest valid pairs can be identified. The problem emphasizes the relationship between the two numbers while adhering to the constraints of digit sums. Ultimately, the solution involves careful calculation and verification of the conditions set forth.
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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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