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What is the sum of the two digit numbers that leave a remainder of 1

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Abhi077
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What is the sum of the two digit numbers that leave a remainder of 1 [#permalink] New post   11 Nov 2018, 09:53
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What is the sum of the two digit numbers that leave a remainder of 1 when divided by both 3 & 4
A)430
B)440
C)450
D)460
E)470
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B
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KarishmaB
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Re: What is the sum of the two digit numbers that leave a remainder of 1 [#permalink] New post   12 Nov 2018, 00:18
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Abhi077 wrote:
What is the sum of the two digit numbers that leave a remainder of 1 when divided by both 3 & 4
A)430
B)440
C)450
D)460
E)470


When the number is divided by 3 or by 4, the remainder is 1.
This means the number is of the form

N = 12a + 1
Since 12 is the LCM of 3 and 4.

All such 2 digit numbers are 12*1 + 1, 12*2 + 1, .... 12*8+1

This is an arithmetic progression of 8 terms with first term as 13 and last term as 97.

Sum = n*(First term + Last term)/2 = 8*(13 + 97)/2 = 440

Answer (B)
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What is the sum of the two digit numbers that leave a remainder of 1 [#permalink] New post   Updated on: 24 Nov 2018, 14:19
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Abhi077 wrote:
What is the sum of the two digit numbers that leave a remainder of 1 when divided by both 3 & 4
A)430
B)440
C)450
D)460
E)470


least two digit number is 3*4+1=13
greatest two digit number is 13+7*12=97
total two digit numbers=7+1=8
13+97=110
110/2=55 mean
55*8=440 sum
B

Originally posted by gracie on 11 Nov 2018, 20:22.
Last edited by gracie on 24 Nov 2018, 14:19, edited 1 time in total.
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Re: What is the sum of the two digit numbers that leave a remainder of 1 [#permalink] New post   24 Nov 2018, 10:42
Abhi077 wrote:
What is the sum of the two digit numbers that leave a remainder of 1 when divided by both 3 & 4
A)430
B)440
C)450
D)460
E)470


Two consecutive numbers never share a common factor other than 1.

Therefore we need LCM of 3 and 4 to find a common number which can be divided by 3 and 4 evenly.

LCM of 3 and 4 = 4*3 = 12

Now, numbers divided by 12 leaving a remainder of 1 are ;

13, 25, 37, 49, 61, 73, 85, 97

Sum (13 + 97) + (25 + 85) + (37 + 73) + (49 + 61)

110 + 110 + 110 + 110 = 440 (B)
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Tiburcio1987
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Re: What is the sum of the two digit numbers that leave a remainder of 1 [#permalink] New post   07 Sep 2020, 03:02
Thank you, can you please explain the formula to get the total two digit numbers? How do you get 7+1=8?
All the best

gracie wrote:
Abhi077 wrote:
What is the sum of the two digit numbers that leave a remainder of 1 when divided by both 3 & 4
A)430
B)440
C)450
D)460
E)470


least two digit number is 3*4+1=13
greatest two digit number is 13+7*12=97
total two digit numbers=7+1=8
13+97=110
110/2=55 mean
55*8=440 sum
B
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What is the sum of the two digit numbers that leave a remainder of 1 [#permalink] New post   27 Dec 2023, 03:35
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What is the sum of the two digit numbers that leave a remainder of 1 when divided by both 3 & 4?

Two digit numbers which will satisfy the condition will be 1 + Multiple of LCM(3,4)
= 1 + 12k (where k is an integer)

So, Series will be 13, 25, ..., 97 (As 97 = 12*8 + 1)

Arithmetic series with first term as 13, Last term as 97, Common difference d = 12 and
Number of terms, n as 8 (as 13= 12*1 + 1 and 97 = 12*8+1)

[We can also sue below formula to find n
(Last Term - First term)/d + 1 = \(\frac{97 - 13}{12}\) + 1 = 7 + 1 = 8]

=> Sum = n * (First Term + Last Term)/2 = 8 * \(\frac{(13 + 97)}{2}\) = 440

So, Answer will be B
Hope it helps!

Watch the following video to MASTER Sequence problems

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What is the sum of the two digit numbers that leave a remainder of 1 [#permalink]
27 Dec 2023, 03:35
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