Its here too!

This is very similar to fist problem of British Mathematical Olympiad 2002-3

This was actually very similar to the recent AMC problem...

From where u got a+b+1=0( mod 9) and 4-a+b=0 (mod 11)?plz explain the solution in details.

SOLUTION We know that a number is divisible by $9$ if and only if the digits sum to a multiple of $9$. Since the digits sum to $136+a+b$, $136+a+b \equiv 8 \pmod9 \implies 1+a+b \equiv 0 \pmod9$. Furthermore, we know that a number is a multiple of $11$ if and only if the sum of the digits of all numbers in odd places minus the sum of the digits of all numbers in even place is divisible by $11$. Mere computation yields that the sum of the numbers in odd places is $77+a$ and the sum of the numbers in even places is $59+b$. Hence, $18+a-b \equiv 0\pmod{11} \implies a-b \equiv 4\pmod{11}$. This means that $a-b=4$ or $b-a=7$. From our first result, $1+a+b \equiv 8 \pmod9$, so $a+b=17,8$. If $a+b=17$, then $a=8,b=9$ or $b=8,a=9$. This clearly does not satisfy $a-b \equiv 4\pmod{11}$. Hence, $a+b=8$. If $a-b=4$, then $a=6,b=2$, which does indeed work. If $b-a=7$, then $a=\frac{1}{2},b=\frac{15}{2}$, which doesn't work given our condition that $a,b$ are integers. Hence, $(a,b)=\boxed{(6,2)}$.

IstekOlympiadTeam wrote: In decimal representation $$\text {34!=295232799039a041408476186096435b0000000}.$$Find the numbers $a$ and $b$. mod 9 and mod 11

Let's get the sum of the numbers except a and b .The Sum of the numbers is 132 and 34! can be divided by 9 but 132 can't be divided there is two option it must 135 or 144 then ,in first option a+b=3 in 2. option a+b=12.Now lets look for it's divisibility for 11

Mod 11 and Mod 9 should do the trick