How many positive two-digit integers have a remainder of 1 when divide

Question

How many positive two-digit integers have a remainder of 1 when divided by 2, a remainder of 2 when divided by 3, and a remainder of 4 when divided by 5?

A. 1
B. 2
C. 3
D. 4
E. 5

Crater · Accepted Answer

How many positive two-digit integers have a remainder of 1 when divided by 2, a remainder of 2 when divided by 3, and a remainder of 4 when divided by 5?

The concept is very simple: If difference between divisor and remainder is constant, take LCM and subtract 1 from it.
eg, 
Divisor - 2,3,5
Remainder - 1,2,4
2-1=3-2=5-4

In this case, LCM of 2,3,5 = 30
Common difference = 1
Total two digit numbers possible, 29, 59, 89

Option (C)

Can,Will · Answer

Answer should be C. 
LCM of 2,3and5 is 30. 
So we have three 2 digit numbers i.e 29,59 and 89 .

Posted from my mobile device

BrushMyQuant · Answer

A remainder of 1 when divided by 2

=> We are looking for odd numbers

A remainder of 2 when divided by 3

=> Number, n = 3k + 2 (where k is an integer)

A remainder of 4 when divided by 5

=> Number, n = 5t + 4 (where t is an integer)

=> n = 3k + 2 = 5t + 4
=> 5t = 3k - 2
=> t =  \frac{3k - 2}{5}

Now, only those non-negative values of k which will give t also as an integer and give odd values of n are possible

We looking for multiples of 3 which have 2 or 7 in the units' digit so that 2-2 or 7-2 will be a multiple of 5, but we also looking at odd values of n 
 If 3k is even then n = 3k + 2 = even, which is not possible as we need odd number 
 So, we need to look for units' digit of 7 for multiple of 3

=> k = 9 => as 3k = 27 (units' digit of 7) => n = 3*9 + 2 = 29
=> k = 19 => as 3k = 57 (units' digit of 7) => n = 3*19 + 2 = 59
=> k = 29 => as 3k = 87 (units' digit of 7) => n = 3*29 + 2 = 89
Looking at the pattern next value of n will be 89 + 30 = 129 which will be a 3 digit number

=> 3 values are possible.

So,  Answer will be C 
Hope it Helps!

Watch following video to MASTER Remainders

https://www.youtube.com/watch?v=P0S6j2uTaj4

saynchalk · Answer

Bunuel

Could you let me know if this makes sense or is it conceptually unsound?

For 1st case (Remainder 1 when divided by 2)

90/2 = 45 numbers

For 2nd Case (R 2 divided by 3)

First number for this case = 11
Last number = 98

Number of digits of this sort = 30

Similarly for 3rd Case --> We have 18 numbers of this sort

Total types of numbers of this kind

45 - 30 - 18 = -3 but i took the absolute value as number of digits can't be negative.

ivycation2024 · Answer

So we take LCM and subtract the common difference from it, or subtract 1 from it? Love this approach btw, very logical, very short and very GMAT!

soniasw16 · Answer

What is the formula/concept behind this?

svsivakrishna · Answer

I guess you can look it this way:

Given: Positive 2 digit number

a) R emainder of 1 when divided by 2

we can consider the number of the form 2N-1

b) a remainder of 2 when divided by 3,

we can consider the number of the form 3N-1

c) a remainder of 4 when divided by 5?

we can consider the number of the form 5N-1

Since 2,3 and 5 all are primes, the LCM is 30

therefore the number is of the form 30N-1 to satisfy all three conditions

so for N=1, number becomes 29 
  for N=2, number becomes 59 
 & for N=3, number becomes 89

beyond that , the number becomes 3 digit number.

Therefore the answer is  3- Option C

How many positive two-digit integers have a remainder of 1 when divide

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

How many positive two-digit integers have a remainder of 1 when divide

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards