Question

there are 12 two digit numbers leave the remainder when digit by 7 the misal digit of a 3 digit numbers is the half the sum of other two digit number the number is 7 more than 35 time the sum of it's extrem digit the new number of 10 by reversing there digit is 396 more than the original numbers find the original numbers​​

Answer 1

Least 3 digit number is 100 ,

Since divisor is 9, & remainder =2

So, 9 x 11 + 2 = 101 is the first 3 digit number which when divided by 9, leaves remainder =2

Next no will be 101+9 = 110, & next = 110+9 =119

& the last 3 digit no leaving remainder 2, while dividing by 9 is 992

So we get an AP series

101, 110, 119, 128 ………………992

Here, Tn = a + (n-1) *d = 992 ,

Where a= 1st term, d= common difference, & n is the term

=> 101 + (n-1)*9 = 992

=> 9n - 9 = 891

=> 9n = 900

=> n = 100

So, there are 100 such numbers ………..ANS

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First number =101

And last numbet = 992

So by last term of A.P = a+(n-1)d

992= 101+(n-1)*9

891/9 = n-1

99= n-1

So n = 100

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1.2k Views ·

This is a question of Arithmatic Progression.

First term: 101

Last term: 992

Common Difference: 9

So, let there are n terms in this series.

So, 992=101+(n−1)×9

or, [8919]=(n−1)

or, n = 99 + 1 = 100 (Answer)

1k Views · Answer requested by

Using divisibility rule of 9 ie no is divisible when sum of digit is nine first such no after 100 is 108. Now go on adding 9 subsequently we get

108,117,126,135,144,153,162,171,180,189,198 means 11 nos between 100 and 200.

207,216,225,234,243,252,261,270,279,288,297

Means 11 nos between 200 and 300. So between 100 to 999, we have 9 such series and each series has 11 such numbers.

So total 99 such no divisible 9 add 2 to each to get 2 as remonder. However biggest number 999 cannot be taken as adding 2 will convert to 4 digit one. So answer is 98.