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Re: How many two-digit numbers are there whose remainder when divided by 1
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31 Jan 2016, 11:46
2 digit numbers whose remainder when divided by 10 is 1 are 11 , 21 , 31 , 41 , 51 , 61 , 71 , 81 , 91 Out of above , numbers whose remainder when divided by 6 is 5 are 11 , 41 and 71
Answer A
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Re: How many two-digit numbers are there whose remainder when divided by 1
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28 Aug 2018, 12:42
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Bunuel wrote:
How many two-digit numbers are there whose remainder when divided by 10 is 1, and whose remainder when divided by 6 is 5?
A. 3 B. 4 C. 5 D. 6 E. 7
IMPORTANT: When it comes to remainders, we have a nice rule that says: If N divided by D leaves remainder R, then the possible values of N are R, R+D, R+2D, R+3D,. . . etc. For example, if k divided by 5 leaves a remainder of 1, then the possible values of k are: 1, 1+5, 1+(2)(5), 1+(3)(5), 1+(4)(5), . . . etc.
Let's start with the second piece of information... The remainder is 5 when the number is divided by 6 So, the possible values are: 5, 11, 17, 23, 29, 35, 41, 47, 53, 59, 65, 71, 77, 83, 89, 95,... (we'll stop here, since the questions specifies that the number is a 2-digit number)
The remainder is 1 when the number is divided by 10 So, the possible values are: 1, 11, 21, 31,... Notice this this basically tells us that the number has a UNITS DIGIT of 1
So, go back to examine the first set of possible values: {5, 11, 17, 23, 29, 35, 41, 47, 53, 59, 65, 71, 77, 83, 89, 95} 3 values have a UNITS DIGIT of 1
So, there are 3 values that satisfy both conditions. Answer: A
How many two-digit numbers are there whose remainder when divided by 1
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Updated on: 25 Oct 2019, 09:37
1
Bunuel wrote:
How many two-digit numbers are there whose remainder when divided by 10 is 1, and whose remainder when divided by 6 is 5?
A. 3 B. 4 C. 5 D. 6 E. 7
Of all the 2-digit numbers that have a remainder of 1 when divided by 10, only 11, 11 + LCM(6, 10) = 11 + 30 = 41, and 41 + 30 = 71, when divided by 6, yield a remainder of 5.
Re: How many two-digit numbers are there whose remainder when divided by 1
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22 Oct 2019, 22:45
ScottTargetTestPrep wrote:
Bunuel wrote:
How many two-digit numbers are there whose remainder when divided by 10 is 1, and whose remainder when divided by 6 is 5?
A. 3 B. 4 C. 5 D. 6 E. 7
Of all the 2-digit numbers that have a remainder of 1 when divided by 10, only 11, 11 + GCF(6, 10) = 11 + 30 = 41, and 41 + 30 = 71, when divided by 6, yield a remainder of 5.
Answer: A
Isn’t GCF(6,10) 2 how did you get to 30?
I tried the n=24x + …. But I can’t find the value of … Could someone please advise? Thanks
How many two-digit numbers are there whose remainder when divided by 1
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25 Oct 2019, 09:38
David nguyen wrote:
ScottTargetTestPrep wrote:
Bunuel wrote:
How many two-digit numbers are there whose remainder when divided by 10 is 1, and whose remainder when divided by 6 is 5?
A. 3 B. 4 C. 5 D. 6 E. 7
Of all the 2-digit numbers that have a remainder of 1 when divided by 10, only 11, 11 + GCF(6, 10) = 11 + 30 = 41, and 41 + 30 = 71, when divided by 6, yield a remainder of 5.
Answer: A
Isn’t GCF(6,10) 2 how did you get to 30?
I tried the n=24x + …. But I can’t find the value of … Could someone please advise? Thanks
We actually meant LCM(6, 10) instead of GCD(6, 10); you're absolutely right that GCD(6, 10) is 2. Thanks for pointing it out.
If the "n" in your equation represents a two digit number that produces a remainder of 1 when divided by 10 and a remainder of 5 when divided by 6; I don't think you'll be able to find the numbers using any equation of the sort n = 24x + ... You can proceed as follows in order to find the numbers that way:
Since n produces a remainder of 1 when divided by 10, n = 10k + 1 for some k.
Since n produces a remainder of 5 when divided by 6, n = 6s + 5 for some s.
Let's add 19 to each equation:
n + 19 = 10k + 20
n + 19 = 6s + 24
Notice that 10k + 20 is divisible by 10 and 6s + 24 is divisible by 6. Since n + 19 is both divisible by 10 and by 6, n + 19 must be divisible by the LCM of 10 and 6, which is 30. Thus, the smallest possible value of n + 19 is 30; which yields the smallest possible value of n is 30 - 19 = 11.
Once we find the smallest possible value of n, we can just add LCM of 6 and 10 to find other two digit numbers which satisfy the requirements of the question; the numbers are 11, 11 + 30 = 41 and 41 + 30 = 71.
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