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Bunuel
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2 digit numbers that are multiples of 14 are 14,28,42,56,70,84,98. So, there are 7 numbers, viz., 15, 29, 43, 57, 71, 85, 99 that would give a remainder of 1 when divided by 14.

Multiples of 14 that are also multiples of 4 are 28, 56, 84. So, there are 3 such numbers, viz., 29, 57, and 85.

Ans (D).
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Bunuel
How many two-digit numbers yield a remainder of 1 when divided by both 4 and 14?

A. 0
B. 1
C. 2
D. 3
E. 4


Kudos for a correct solution.

The correct method would be to find LCM and then add 1 to {LCM, 2*LCM, and so on}
LCM of 4 & 14 = 28..
first required number=29, next 2*28+1 or 57 and third 3*28+1=85, but thereafter, it will be a 3 digit number as 4*28+1=113

So ans 3: 29, 57 and 85

D
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Bunuel
How many two-digit numbers yield a remainder of 1 when divided by both 4 and 14?

A. 0
B. 1
C. 2
D. 3
E. 4


Kudos for a correct solution.

LCM of 4 & 14 is 28, hence 1st 2-digit no. which yields rem. of 1 when divided by both 4 & 14 is 28+1 = 29
2nd no. would be 28*2 +1 = 57

this is of the form 28k+1, where k is an integer and k>0

hence,
for k = 1, 28k+1 = 29
k = 2, = 57
k = 3, = 85
k = 4, = 113......we will stop here as this is now 3 digits,

hence total 2-digit nos. are 3

And. D
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Bunuel
How many two-digit numbers yield a remainder of 1 when divided by both 4 and 14?

A. 0
B. 1
C. 2
D. 3
E. 4


Kudos for a correct solution.

800score Official Solution:

Let’s use n to denote a two-digit number that fits the requirement in the question.

Since we are looking for a remainder of 1 when n is divided by 4 or 14, then (n – 1) must be divisible by both 4 and 14. All numbers divisible by both 4 and 14 must be divisible by their least common multiple, which is 28.

So n – 1 can equal any two-digit multiple of 28. These possible values are:
n – 1 = 28, 56, or 84.

Therefore, n = 29, 57, or 85 (3 different two-digit numbers).

The correct answer is choice (D).
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Bunuel
Bunuel
How many two-digit numbers yield a remainder of 1 when divided by both 4 and 14?

A. 0
B. 1
C. 2
D. 3
E. 4


Kudos for a correct solution.

800score Official Solution:

Let’s use n to denote a two-digit number that fits the requirement in the question.

Since we are looking for a remainder of 1 when n is divided by 4 or 14, then (n – 1) must be divisible by both 4 and 14. All numbers divisible by both 4 and 14 must be divisible by their least common multiple, which is 28.

So n – 1 can equal any two-digit multiple of 28. These possible values are:
n – 1 = 28, 56, or 84.

Therefore, n = 29, 57, or 85 (3 different two-digit numbers).

The correct answer is choice (D).

Thanks Bunuel, this is more smarter way!
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LCM (4, 14) =28
two digit multiple of 28 are 28, 56 and 84
two digits numbers that yield remainder of 1 when divisible by 4 and 14 are 28+1, 56+1 and 84+1
ans: D
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JJo
Bunuel
How many two-digit numbers yield a remainder of 1 when divided by both 4 and 14?

A. 0
B. 1
C. 2
D. 3
E. 4


Kudos for a correct solution.

LCM of 4 & 14 is 28, hence 1st 2-digit no. which yields rem. of 1 when divided by both 4 & 14 is 28+1 = 29
2nd no. would be 28*2 +1 = 57

this is of the form 28k+1, where k is an integer and k>0

hence,
for k = 1, 28k+1 = 29
k = 2, = 57
k = 3, = 85
k = 4, = 113......we will stop here as this is now 3 digits,

hence total 2-digit nos. are 3

And. D

Could you explain further why you add one? I did this y=28x + 29… but I can’t do it quicker without finding the first common term in the traditional way. Thanks
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