Number Theory - Solutions to final few questions

Have given below the solutions to the final few questions on Number Theory. The questions can be found here .
1. A 4-digit number of the form aabb is a perfect square. What is the value of a-b?

A number of the form aabb has to be a multiple of 11. So, it is the square of either 11 or 22 or 33 or...so on up to 99.

88^2 = 7744. This is the only solution possible. Most of these trial and error questions need to be narrowed down a little bit before we can look for the solution. That narrowing down is critical. In this case, we should look for multiples of 11.

2. LCM of three numbers is equal to 1080; HCF of the three numbers is equal to 2. How many such triplets are possible?

HCF = 2. Let the numbers be 2x, 2y and 2z. HCF of (x,y,z) =1
LCM = 2xyz. => xyz = 540
540 = 22 * 33 * 5

x = 22 , y = 33, z = 5
x = 22 * 33, y = 5, z = 1
x = 22 * 5, y = 33, z = 1
x = 33 * 5, y = 22, z = 1
x = 22 * 33 * 5, y=1, z=1

These are the basic combinations. The first 4 yield 6 ordered triplets each, the last one yields 3 ordered triplets.

So, there are 5 unordered triplets and 27 ordered triplets

3. N leaves a remainder of 4 when divided by 33, what are the possible remainders when N is divided by 55?

The LCM of 33 and 55 is 165. This is the starting point. If a number leaves a remainder of 4 when divided by 33, it can leave remainders 4, 37, 70, 103, and 136.

Or the number can be of the form 165n +4, or 165n + 37 or 165n + 70,103 or 136

When divided by 55, the possible remainders are 4, 37, 15, 48 and 26


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IIM CAT Preparation Tips: Number Theory - Solutions to final few questions

Nov 15, 2010

Number Theory - Solutions to final few questions

Have given below the solutions to the final few questions on Number Theory. The questions can be found here .
1. A 4-digit number of the form aabb is a perfect square. What is the value of a-b?

A number of the form aabb has to be a multiple of 11. So, it is the square of either 11 or 22 or 33 or...so on up to 99.

88^2 = 7744. This is the only solution possible. Most of these trial and error questions need to be narrowed down a little bit before we can look for the solution. That narrowing down is critical. In this case, we should look for multiples of 11.

2. LCM of three numbers is equal to 1080; HCF of the three numbers is equal to 2. How many such triplets are possible?

HCF = 2. Let the numbers be 2x, 2y and 2z. HCF of (x,y,z) =1
LCM = 2xyz. => xyz = 540
540 = 22 * 33 * 5

x = 22 , y = 33, z = 5
x = 22 * 33, y = 5, z = 1
x = 22 * 5, y = 33, z = 1
x = 33 * 5, y = 22, z = 1
x = 22 * 33 * 5, y=1, z=1

These are the basic combinations. The first 4 yield 6 ordered triplets each, the last one yields 3 ordered triplets.

So, there are 5 unordered triplets and 27 ordered triplets

3. N leaves a remainder of 4 when divided by 33, what are the possible remainders when N is divided by 55?

The LCM of 33 and 55 is 165. This is the starting point. If a number leaves a remainder of 4 when divided by 33, it can leave remainders 4, 37, 70, 103, and 136.

Or the number can be of the form 165n +4, or 165n + 37 or 165n + 70,103 or 136

When divided by 55, the possible remainders are 4, 37, 15, 48 and 26


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