Subject: Curious student

Hello, my name is Wallace. I'm a highschool student. The difficulty level of this question is probably grade 12.

Here's the question:

A six-digit integer XYXYXY, where X and Y are digits is equal to five times the product of three consecutive odd integers. Determine these three odd integers.

Thank you, your help is appreciated.

Hi Wallace,

Expanding the six digit number into place value notation gives:

 XYXYXY = X(105) + Y(104) + X(103) + Y(102) + X(101) + Y(100) = X(100000) + X(1000) + X(10) + Y(10000) + Y(100) + Y(1) = X(100000 + 1000 + 10) + Y(10000 _ 100 + 1) = 101010X + 10101Y = 10101(10X + Y) = 3x7x13x37(10X + Y)

XYXYXY is 5 times some number so XYXYXY is a multiple of 5 This means the unit digit of XYXYXY must be 0 or 5 Thus, Y=0 or 5

Suppose Y=0 Then XYXYXY = 3x7x13x37(10X). The factor 10 in the factorization of XYXYXY means XYXYXY must be EVEN. But 5 times the product of three odd integers is ODD. So we have a contradiction and can't have Y=0 and thus Y=5.

Using the factorization again:

XYXYXY = 3x7x13x37(10X + 5) = 3x7x13x37x5(2X + 1)

This factorization is 5 times the product of three consecutive odd integers so, by cancelling the factor 5, we have that: 3x7x13x37(2X + 1) is the product of three consecutive odd integers. But 3x7x13x37(2X + 1) = 37x39x7(2X+1) so, 7(2X+1) = 35 or 7(2X+1) = 41. The first equation gives X=2. The second equation has no integer solution. Therefore X=2 the six digit number is 252525.

Check: 35x37x39x5 = 252525

Cheers,
Paul
Go to Math Central