# Thread: need help for my math problem

1. ## need help for my math problem

A person went to a bank to cash a check. In handing over the money the cashier, by mistake, gave him dollars for cents and cents for dollars. He kept the money in his pocket without examining it, and spent a nickel on his way home. He then found that he has exactly twice the amount of the check. He had no money in his pocket before going to the bank. What was the exact amount of that check?

Please give me the solution for this puzzle and tell me how to solve it also?

Thanks

2. ## Re: need help for my math problem

Let C = number of cents on the check and D = number of dollars. The amount of money he should have received in cents is 100D + C, but he actually receved 100C + D. So after spending the nickel and realizing he has twice as much money as he should have the equation you need to solve is:

100C + D - 5 = 2(100D +C)

Rearrange: 98C = 199D + 5

Both C and D must be integers, and both must be in the range (0,99) inclusive. At this point you can use trial and error, but first note that the left side is even, which means 199D must be odd, so D is odd. Also, since 98 has prime factors 2 and 7, D cannot be a multiple of 7 (since 199D +5 would not be a multiple of 7). After that, I think it's just trial and error to find integer solutions for C and D, but to save you a little time I suggest you start with trying D = 25, then 27, etc. - you should get it shortly

3. ## Re: need help for my math problem

Originally Posted by ebaines
Let C = number of cents on the check and D = number of dollars. The amount of money he should have received in cents is 100D + C, but he actually receved 100C + D. So after spending the nickel and realizing he has twice as much money as he should have the equation you need to solve is:

100C + D - 5 = 2(100D +C)

Rearrange: 98C = 199D + 5

Both C and D must be integers, and both must be in the range (0,99) inclusive. At this point you can use trial and error, but first note that the left side is even, which means 199D must be odd, so D is odd. Also, since 98 has prime factors 2 and 7, D cannot be a multiple of 7 (since 199D +5 would not be a multiple of 7). After that, I think it's just trial and error to find integer solutions for C and D, but to save you a little time I suggest you start with trying D = 25, then 27, etc. - you should get it shortly
Thank you for reply