Question 416055: I am a seven-digit number. My first five digits are 2, 1, 3, 5, and 8. I am divisible by 99. What are my last two digits?
I have tried 11, 23, 22, 25, 24, 69. I can't figure it out, will you help me?
Answer by richard1234(7193)   (Show Source):
You can put this solution on YOUR website! We can write the number in the form 21358ab, where a,b are digits.
We know that the sum of the digits is a multiple of 9, so 2+1+3+5+8+a+b ≡ 0 (mod 9) --> 19+a+b ≡ 0 (mod 9) --> a+b ≡ 8 (mod 9), a+b = 8 or 17.
An integer is a multiple of 11 if and only if the difference between the odd-positioned digits and the even-positioned digits is a multiple of 11. Hence, (2+3+8+b) - (1+5+a) ≡ 0 (mod 11), 7+(b-a) ≡ 0 (mod 11), b-a ≡ 4 (mod 11), a-b = 7 or -4.
We have the system of equations a+b = 8, 17 and a-b = 7, -4. If we add the equations, we expect an even number, so the only possible ordered pairs(a+b, a-b) are (8, -4) and (17, 7). Do each case separately:
a+b = 8
a-b = -4
This implies a = 2 and b = 6.
a+b = 17
a-b = 7
This implies a = 12, impossible.
The only possible case is a = 2, b = 6. It can be checked that 2135826/99 = 21574.
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