NMTC Practice Part 2

First, let see $a,b,c\le 9$ . Now, The equation $64a+8b+c=403$ can be factorized as $8(8a+b)+c=403$ . So, if $c\le 9$ then the only solution is $c=3$ and $8a+b=50$ . Remember the first inequality $a,b,c\le 9$ then the only solution is $a=6,b=2$ .

Finally, the number $\overline { abc } =\boxed { 623 }$ .

Because 64a+8b+c=403 the maximum value of "a" is 6.Now let us put in 6 in place of "a" and we get that 8b+c=19.Now,"b" can't be 1 but it can be 2,putting in 2 in place of "b" gives c=3.BINGO!!

Recall that $\overline{abc}=100a+10b+c$ . Looking at the expression $64a+8b+c$ can give you a hint that we're looking for a base-8 expansion of some number. Apparently, at this point $403$ is written in base-10. Fetch your calculator (or use brains) to obtain the conversion $403_{10} = 623_8$ . Thus, going digit-by-digit,

$a=6 \\ b=2 \\ c=3$

Max value of a=6, putting 6 in equation, we get b=2 and c=3

I just tried the trial and error method .. First, I started with the maximum value of A -- which is 6 .. Then I solved the values of b and c, which is 2 and 3 respectively.. Lol I just solved it for 1 minute.. Trial and Error method is sometimes more easier for me

when you try the 6,2,4 the answer is 403

(A off-topic question) Can someone please tell me how to participate in NMTC?

work mod 8 to conclude that c=3 work mod 8 again to conclude that b=2