Question

When a number is divided by 17, the remainder is 9. When the same number is divided by 23, the remainder is 4. Find the number?​

Answer 1

Answer:

Let Z be the number

Z = 13x + 11 where x is the quotient when Z is divided by 13

Z = 17y + 9 where y is the quotient when Z is divided by 17

13x + 11 = 17y + 9

13x + 2 = 17y since x and y are quotients they should be whole numbers . Since y has to be a whole number the left hand side should be multiple of 17

The least possible value of x satisfying the condition is 9 and y will be 7

The answer is 13*9 + 11 = 128 or it is 17*7 + 9 = 128

This is the least number possible. There will be multiple answers and will increase in multiples 17*13 = 221 like 349 , 570, etc

Step-by-step explanation:

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Answer 2

The required number is 349.

Given:

Dividend = x

Quotient 1  = q1

Quotient 2  = q2

Remainder 1 = 9

Remainder 2 = 4

Divisor 1 = 17

Divisor 2 = 23

To Find:

Find the number x.

Solution:

Let x be the number.

Dividend = Divisor * Quotient + Remainder

According to the given conditions, we get two equations

x = 17q1 + 9 ... eqn (1)

x = 23q2 + 4 ... eqn (2)

Equating equations (1) and (2) we get,

13q2 - 17q1 = 9 - 4

23q2  -17q1 = 5

$q2 = \frac{17q1 + 5}{23}$

The least value of p for which q is a whole number is p = 20.

Substitute the value of p in eqn 1,

x = 17(20) + 9

x = 349

Hence, the number is 349.