How many four digit numbers are there such that when they are divided by 101 they have 99 as remainder?

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Solution

We observe that 1008 is the least number of 4 digits which gives a remainder 99 when divided by 101 and 9997 is the greatest.

So the numbers are 1008, 1109, 1210, 1311,……………..9997. These are in Arithmetic sequence in which first term a = 1008, common difference

d = 101 and the last term l= 9997. We use the formula l = a + (n-1)d to find n.

So, 9997 = 1008 + (n-1) 101 or 9997 - 1008 = (n-1) 101 or 8989 = (n-1)101. So, n-1 = 8989/101 = 89. This gives n = 90.

Therefore, there are 90 four-digit numbers which leave remainder 99 when divided by 101.

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