Remainder obtained when a no. is divided by a divisor is 13. When no. is tripled & divided by same divisor,?

Question

remainder obtained is 1. Which could be the divisor among- 19/47/43/29.

I want "general strategy/method" to solve these type of questions. -thanks.

Answer 1

The strategy is: Forget the number and concentrate on the remainder.

n / x = k, R13
3n / x = 3k, R39

So how do you get from 3*13 = 39 for a remainder to having a remainder of only 1?

It must be that 39 divided by the divisor gives a remainder of 1. The only divisor shown that works is 19.

(While 39 / 2 also gives a remainder of 1 in the second case, dividing by 2 wouldn't give a remainder of 13 in the original case.)

Answer 2

Let n be the number and d the divisor. Then,

n = q * d + 13, where q is the quotient of the division. It follows that

3n = (3q) * d + 39

If d > 39, then the remainder is 39

If d <= 39, then there exists and integer r, 0 <= r <= 38, such that 39 = d + r. So,
3n = (3q) * d + d + r = (3q + 1) d + r. Since 0 <= r < d, it follows the remainder is r = 39 -d.

Conclusion.

r = 39 - d ,if d <= 39.
r = 39 if, d > 39

This same process can be used for numbers other than 13.

Answer 3

19

Let N be the number and D be the devisor

N = KD + 13
=> 3N = 3KD + 39

Now if we divide this by D then of course 3KD will not give any remainder because it is divisible by D
So the remainder of 3N/D will be the same as the remainder of 39/D (which is 1 (given))

=> 39 = X*D + 1 where X is any number

only 19 fits the criteria (39 = 2*19 + 1)

Answer 4

Let:
n be the given number to be divided
x to be the divisor
k to be a constant (whole number)
n/x = k + 13/x
3n/x = 3k + 39/x, 39/x >1
39/x = m + 1/x , where m is a whole number.
39 = mx +1
38 = mx
Therefore, as m is a whole number, x must be either divisible by 2 or 19. SO the answer is 19.

Answer 5

This is called modulo arithmetic.

x mod y = 13

3 * x mod y = 1

It is a Boolean Algegra question... And you want to know y, and not x. Interesting...

And it is interesting that he didn't want the answer, but the general solution method to the problem. That is going to require a Boolean solution, as I said... The general case cannot be stated in algebra because there are too many unknowns, but the "mod" operation reduces this to a simultaneous linear equation in two variables. The trick is knowing the "mod" identities.

I don't, sorry...

Ron.