maths question?

Question

Find a whole number so that:

A=13, y=2

When you divide this number by 131 the remainder is Y,
When you divide this number by A the remainder is 7.

Answer 1

Simultaneous equations

Let n = unknown number

Eq 1 becomes {(n/131) +2 = 0}
Eq 2 becomes {(n/ 13 ) + 7 = 0}

Expand and collect terms - simplify - substitute and solve...

Hope this helps :-)

Answer 2

Ok, not too tough this time.

You need a number that's 7 plus a multiple of 13, yet also 2 plus a multiple of 131.

Let's look at the second part first:

Possible numbers are 2, 133, 264, 395, 526, 657, 788 etc.

Now we check each of these numbers for their remainders after subtraction of multiples of 13

We see 2, 3, 4, and so forth. In short, when the multiple of 131 in the first number is 0, the remainder after this second test is 2. This remainder goes up by 1 for each multiple of 131 added in the initial sequence.

Since the remainder we wish to see is 7, i.e. 5 higher than the original 2, we need to add 5 'lots' of 131 to the original number 2 from that first sequence.

5 x 131 = 655, 2 + 655 = 657, and so the first available number is 657.

Since the test for the remainder relies on subtraction of multiples of 13, adding multiples of 13x131=1703 will result in qualifying numbers also.

Therefore you can use 657 with the addition of any multiple of 1703 (including zero, of course) to achieve your aim.

I hope this helps, but please feel free to drop me a line if you'd like to go over this or any other figure-work in further detail.

Answer 3

N=131*x + 2
N= 13*y + 7
131*x = 13*y + 5, need to choose any integer x,y that solve this.
Picking y=10*x looks promising to give an integer sol'n for x ...
131*x = 130*x + 5
x=5, y=50
N = 657

Answer 4

657

Answer 5

657

Answer 6

The answer is 657

Answer 7

657 is the answer