Find the sum of all two digit greater than 10 which when divided by 5 leaves 4
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Step-by-step explanation:
Let x is at 10’s place and y at 1′ place
then 10x+y=5n+1
y=5n-10x+1
y=5(n-2x)+1
As y is positive number
so y=5q+1 ( where q=n-2x is a positive integer)
Now max value of y is 9
so q=0 or 1
and y=1 or 6
Thus all the 2 digit numbers with 1 or 6 at 1’s place will satisfy the situation
S no are 11,21,31…………………………………91
and 16,26,……………………………………………..96
Total 18 numbers are possible
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