CAT#9:What is the value of the two-digit positive integer n?

[ssr174]

What is the value of the two-digit positive integer n?

(1) When n is divided by 5, the remainder is equal to the tens digit of n.

(2) When n is divided by 9, the remainder is equal to the tens digit of n.

I read the explanation to this question and I'm not sure I understand how the two lists for each statement are generated. Why is 49 the overlap? Is there an alternate way to approaching this problem?

[vikash.121186]

What is the value of the two-digit positive integer n?

(1) When n is divided by 5, the remainder is equal to the tens digit of n.

(2) When n is divided by 9, the remainder is equal to the tens digit of n.

I read the explanation to this question and I'm not sure I understand how the two lists for each statement are generated. Why is 49 the overlap? Is there an alternate way to approaching this problem?

***I don't have the explanation but i can try.Hope it helps***
Case 1 : According to the first statement, the remainder on dividing by 5 is equal to the tens digit of n.

So, let the 2-digit number be xy which is nothing but 10x+y.
10x is divisible by 5 and thus the remainder will come from y.
In order for y to give x as remainder, y should be either equal to x(and less than 5) or equal to x+5(and x should be less than 5).

For the case,where y is equal to x(and less than 5), we have the following numbers 11,22,33,44.

For the case, where y is equal to x+5(and x is less than 5), we have the following cases 16,27,38,49.

Case 2 : When n is divided by 9, the remainder is tens digit of n.
Any 2 digit number say xy can be written as 9x+x+y. Since 9x is divisible by 9, the remainder shall come from x+y.
For x+y to give remainder as x(x should be less than 9) and y has to be 9 so that the

remainder shall come solely from x or
e.g. of such numbers are 19,29,39,49,59,69,79,89
or
y has to be 0 and x can be any digit less than 9.
e.g of such numbers are 10,20,30,40,50,60,70,80.

The only number that satisfies both the statements is 49.

This is how i approached the question. I am sure there's a better approach to this.

Regards

[jnelson0612]

What is the value of the two-digit positive integer n?

(1) When n is divided by 5, the remainder is equal to the tens digit of n.

(2) When n is divided by 9, the remainder is equal to the tens digit of n.

I read the explanation to this question and I'm not sure I understand how the two lists for each statement are generated. Why is 49 the overlap? Is there an alternate way to approaching this problem?

You know, I would just start listing out values that fit the statements.

1) possible values are 5, 16, 27, 38, 49. I see the pattern that each value increases by 11, and note that you can't have a tens digit beyond 4 if it is a remainder of a number divided by 5. Not sufficient.

2) possible values are 9, 10, 19, 20, 29, 30, 39, 40, 49, etc.

Together: There will only be one value on both lists: 49.

This is old school but I think that it works and gets you somewhere when you may not remember more elaborate math.

[AditiS301]

Hi i came across this post. Was practising Manhattan Mock tests. I answered E. Because 49 and 99 are possible values.

If N =49
N/45 = 45*1 + 4 ( 4 Remainder and tens digit N )

If N =99
N/45 =45*2 + 9 ( 9 remainder and tens digit of N)

I saw the solution, Please guide me why 99 is not considered as a possible value.

[Sage Pearce-Higgins]

Hi AditiS301,

I agree with Jamie on this one: don't try to be too smart with Algebra, just list out the possible values in a simple, logical way. And what's the remainder when 99 is divided by 9? It's not 9, it's 0! So that number doesn't work with statement (2).

[XiaokunY49]

Let's say if this integer is X, n is the remainder after X is devided by both 5 and 9, thus X can be expressed as below:

X=5*a+n & X=9*b+n ==> 5a=9b ==>because X is a 2-digit No, then a=9, b=5,
then X=45+n, because n is equal to the tens digit of X, and X<5(the remainder always < devider),
so, X=45+4=49

Here we have 49!!

[RonPurewal]

Please guide me why 99 is not considered as a possible value.

99 doesn't satisfy statement #1.

in fact, it CAN'T satisfy statement #1. (on division by 5, it's impossible to get a remainder other than 0, 1, 2, 3, or 4.)

[RonPurewal]

...and it doesn't satisfy statement #2, either. (the remainder of 99 ÷ 9 is zero... and the tens digit of 99 is certainly not zero.)

so, i'm not sure where you are getting 99 from—it actually doesn't satisfy EITHER of the two statements!