How many two-digit whole numbers yield a remainder of 1 when divided

n=10p+1 --> Number could be 11 21 31 41 51
n=6q+1 --> Number could be 7 13 19 25 31

n= 30q+31
so n could be 31,61,91

IMO D

Answer = D. Three

LCM of 10 & 6 = 30

Two-digit numbers giving remainder 1 for 30 are

31, 61, 91

Can you explain the highlighted calculation? How is that obtained?

Find the least common factor and multiples of the number +1
Least common factor of 10 and 6 is 30 (two digits multiples of 30 are 30,60,90.. Add +1 to the numbers) so totally 3 numbers are possible

hi Bunuel
don`t you think with respect to your answer..
since q is the quotient..how can you put q=0 and get 1 as common from both equations
I mean if you put q=0,then n=1 but n is a two digit number so the first common value needs to be 31 i.e N(two digit)=30m+31..
thanks

I first found general formula and then applied the restriction.

31,61 & 91 are the only three two digit numbers that when divided by 10 and 6 each leaves a remainder of 1.
Option D.

Hi All,

This type of question is rooted in pattern-matching. Once you find the patterns behind this question, it won't be hard to solve. Instead of trying to do every step all at once, I suggest that you break the prompt into "pieces":

First, name the 2-digit numbers that are evenly divisible by 10:

10, 20, 30, ......90

Now, name the 2-digit numbers that have a remainder of 1 when divided by 10:

11, 21, 31,.....91

Now that we've established the numbers that fit the first 2 "restrictions" in the prompt, let's factor in numbers that are ALSO divisible by 6:

30, 60, 90

And ALSO have a remainder of 1 when divided by 6:

31, 61, 91

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Two digit numbers divided my 10 yielding remainder 1 = 11,21,31,41,51,61,71,81,91

Two digit numbers divided my 6 yielding remainder 1 = 31,61,91

Three common numbers.

Hence (D)

The possible number N can be written as follow:
N = Multiple of LCM(6,10) + 1st such number
N = 30x + 1
Possible values = 1, 31, 61, 91
Answer : 3 such 2 digit number. D.

We have to take the LCM and never the product

N1 = 10 k + 1
N2 = 6k + 1

N3 = 30 k + 1

k = 1,2,3

Only 3, two digit values can be yield.

D

LCM (6,10) = 30
x = 10q+1 = 6r +1
=> x = 30m + 1 -> test m = 1,2,3 -> 31, 61,91

Hi Bunuel
I have a small query regarding this question.

when I tried solving it, I included all positive and negative numbers assuming that the question didn't restrict to positive integers,
and I got -89,-59,-29,31,61,91 (which are six possible two digit numbers)

As the answer is already known, why was my assumption wrong?

LCM(6,10) = 30
Numbers ={31,61,91}

IMO D

Posted from my mobile device

There is an easy way to solve this conceptually.

The only two digit numbers that will produce a remainder of 1 when divided by 10 are two-digit numbers with units digits of 1.
11, 21, 31, 41,51,61,71,81,91

You can quickly go through and determine that only 31,61,91 produce a remainder of 1 when divided by 6 and are two digit numbers.

Two-digit numbers: 10 - 99

Divided by '10' gives remainder '1' and when divided by '6' gives remainder '1'.

That means the number is divisible by the LCM of [10,6] = 30.

Two-digit numbers which will give remainder as '1' on getting divided by '30': 31, 6, 91 = 3

Answer D

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