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06 Dec 2021, 03:45
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If a number when divided by 7 gives a remainder of 4, find the sum of all such two-digit numbers satisfying this condition.
(A) 666
(B) 678
(C) 689
(D) 728
(E) 751
Are You Up For the Challenge: 700 Level Questions
(A) 666
(B) 678
(C) 689
(D) 728
(E) 751
Are You Up For the Challenge: 700 Level Questions
06 Dec 2021, 04:36
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First such number would be 7*1+4 =11
Last number in the sequence 13*7 +4 = 95
AM = 11+95/2 =53
Total terms =13-1 +1 =13
13*53 = 689
Posted from my mobile device
Last number in the sequence 13*7 +4 = 95
AM = 11+95/2 =53
Total terms =13-1 +1 =13
13*53 = 689
Posted from my mobile device
31 Aug 2022, 09:25
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If a number when divided by 7 gives a remainder of 4
Theory: Dividend = Divisor*Quotient + Remainder
Let the number be n
n -> Dividend
7 -> Divisor
a -> Quotient (Assume)
4 -> Remainders
=> n = 7*a + 4 = 7a+4
find the sum of all such two-digit numbers satisfying this condition
n = 7a + 4
To find all the two digit numbers which satisfy this condition we need to put values of a from 1 onwards such that 7a + 4 is a two digit number
a= 1, 7a + 4 = 7*1 + 4 = 11
a= 2, 7a + 4 = 7*2 + 4 = 18
a= 3, 7a + 4 = 7*3 + 4 = 25
and so on...
By now you would have realized that this will form an Arithmetic Sequence with
(Watch this video to Learn About Arithmetic Sequence)
First Term, a = 11
Common Difference, d = 7
Last Term, \(T_n\) = 95 (7*13 + 4)
Let's find how many such 2 digit numbers are there
Number of terms of Arithmetic Sequence (n) is given by ( (Last Term - First Term) / Common Difference) + 1
=> n = \(\frac{95-11}{7}\) + 1 = \(\frac{84}{7}\) + 1 = 12 + 1 = 13
Sum of Arithmetic Sequence, \(S_n\) is given by = Number of Terms * Mean of First and Last Term = 13 * \(\frac{11 + 95}{2}\) = 13 * \(\frac{106}{2}\) = 13 * 53 = 689
So, Answer will be C
Hope it helps!
Watch the following video to learn the Basics of Remainders
Theory: Dividend = Divisor*Quotient + Remainder
Let the number be n
n -> Dividend
7 -> Divisor
a -> Quotient (Assume)
4 -> Remainders
=> n = 7*a + 4 = 7a+4
find the sum of all such two-digit numbers satisfying this condition
n = 7a + 4
To find all the two digit numbers which satisfy this condition we need to put values of a from 1 onwards such that 7a + 4 is a two digit number
a= 1, 7a + 4 = 7*1 + 4 = 11
a= 2, 7a + 4 = 7*2 + 4 = 18
a= 3, 7a + 4 = 7*3 + 4 = 25
and so on...
By now you would have realized that this will form an Arithmetic Sequence with
(Watch this video to Learn About Arithmetic Sequence)
First Term, a = 11
Common Difference, d = 7
Last Term, \(T_n\) = 95 (7*13 + 4)
Let's find how many such 2 digit numbers are there
Number of terms of Arithmetic Sequence (n) is given by ( (Last Term - First Term) / Common Difference) + 1
=> n = \(\frac{95-11}{7}\) + 1 = \(\frac{84}{7}\) + 1 = 12 + 1 = 13
Sum of Arithmetic Sequence, \(S_n\) is given by = Number of Terms * Mean of First and Last Term = 13 * \(\frac{11 + 95}{2}\) = 13 * \(\frac{106}{2}\) = 13 * 53 = 689
So, Answer will be C
Hope it helps!
Watch the following video to learn the Basics of Remainders
