Find the sum of all three digit no, which leaves the remainder 3 when divisivle by 7
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Solution :
Number when divided by 7 will give a remainder of 3. Such numbers will be of the form 7n+3.
Smallest two digit number of the form 7n+3 = 10 for n = 1
Largest two digit number of the form 7n+3 = 94 for n = 13
So, the numbers will be like 10,17,24,……,94.
Total number of terms = 13 [they are in A.P]
Hence , sum of arithmetic progression =(First term + Last term)
=(10 + 94)
=*104
= 13*52 = 676
Hence, sum of all two-digit numbers that give a remainder of 3 when they are divided by 7 = 676
Number when divided by 7 will give a remainder of 3. Such numbers will be of the form 7n+3.
Smallest two digit number of the form 7n+3 = 10 for n = 1
Largest two digit number of the form 7n+3 = 94 for n = 13
So, the numbers will be like 10,17,24,……,94.
Total number of terms = 13 [they are in A.P]
Hence , sum of arithmetic progression =(First term + Last term)
=(10 + 94)
=*104
= 13*52 = 676
Hence, sum of all two-digit numbers that give a remainder of 3 when they are divided by 7 = 676
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