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Let the sum of all two digit numbers which being divided by 4, leave remainder 1 be . Let the sum of all three digit numbers which being divided by 5 , leave remainder 4 be . Find .
The answer is 100480.
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1 solution
We are talking about two arithmetic progression.
case 1, series of two digit numbers such that when divided by 4 leaves 1 as a remainder.
which implies these two digit numbers will satisfy 4a + 1 , where a is an integer.
the first two digit to satisfy 4a + 1 form is 13
4a + 1 >=10 and 4a + 1 <=99
(basically we are looking for multiples of 4 plus 1)
so, the series of numbers will look like 13, 17, 21,....97
nth term = first term + (n-1)*d so, 97 = 13 + (n-1) *4 will give n = 22
N= Sum = (n/2) × (2a + (n-1)d) = (22/2){(2 13) +{ (22-1) 4}] =1210--------result i
similarly, for case 2, series of three digit numbers such that it follows form 5b +4
5b + 4 >=100 and 5b + 4 <= 999 gives the second series... 104, 109,.....999
nth term = first term + (n-1)*d so, 999 = 104 + (n-1) *5 will give n = 180
X = sum = (n/2) × (2a + (n-1)d) = (180/2){(2 104) +{ (180-1) 5}] =99270-
adding N + X = 100480