When a positive integer n is divided by 100, the remainder is the same

Let Q be the quotient as well as the remainder
Therefore the number = Q+100Q = 14K, in which K=positive integers
101Q=14K
LCM of 101 & 14 = 1414
Number of multiples of 1414 between 1000 & 9999 = 7

This is my view on the problem, would request to correct me if i am wrong.

if n/100 quotient=remainder,
then n must take form of 101, 202...1010, 1111...
if n/14 gives 0 remainder,
then n must be multiple of 14
14*101=1414=least value of n
1414+k(14*101)<9999→
1414k<8585→
k<6.1→
k=6
6+1=7 distinct integer values for n
C

n = 100a + a = 101 a
n = 14b
n = 101 a = 14 b
least possible value of n is LCM (101, 14) is 1414.
Multiples of 1414 between 1000 and 9999 are 7

C is correct.

Solution:

If we let the quotient be q when n is divided by 100, we have n = 100q + q (since the remainder is the same as the quotient). If we let the quotient be u when n is divided by 14, we have n = 14u (since the reminder is 0). Therefore, we have:

100q + q = 14u

101q = 14u

Since 101 and 14 are relatively prime, q must be a multiple of 14 and u must be a multiple of 101. Since u is a multiple of a larger number, let’s check the possible values of u and hence the possible values of n (keep in mind that n = 14u and 1000 < n < 9999):

If u = 101, n = 14 x 101 = 1414.

If u = 202, n = 14 x 202 = 2828.

If u = 303, n = 14 x 303 = 4242.

If u = 404, n = 14 x 404 = 5656.

If u = 505, n = 14 x 505 = 7070.

If u = 606, n = 14 x 606 = 8484.

If u = 707, n = 14 x 707 = 9898.

We see that there are 7 possible values of n (notice that if u = 808, n will be greater than 9999).

Answer: C

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Given

• When a positive integer n is divided by 100, the remainder is the same as the quotient.
• When n is divided by 14, the remainder is 0.

To Find

• The number of distinct integer values can n take if n is more than 1000 and less than 9999.

Approach and Working Out

• As the remainder and the quotient are the same, n = 100x + x
• As it is a multiple of 14, 100x + x = 14k

o 101x = 14k = n
• n must be a multiple of 101 and 14 so it must be a multiple of 101 × 14 = 1414.

o Number of multiples of 1414 in the range 1000 to 9999 is,
o = {9999/1414} – {1000/14}
o = 7 – 0 = 7

Correct Answer: Option C
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n = 100 p + p or p( 100 + 1) = 101p

n = 14q

So 'n' is multiple of 101 and 14 [LCM of 101 and 14 is 1414]

=> \(\frac{9999 }{ 1414}\) = 7.xx and hence total numbers greater than 1,000 and less than 9999 are seven.

Answer C

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