### CAT 2021 Crash Course + Mock Test Series (INR 4999 Only)

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**Question 1:**

Three company of soldiers containing 120, 192, and 144 soldiers are to be broken down into smaller groups such that each group contains soldiers from one company only and all the groups have equal number of soldiers. What is the least number of total groups formed?

**Answer & Solution**

The least number of groups will be formed when each group has number of soldiers equal to the HCF. The HCF of 120, 192 and 144 is 24. Therefore, the numbers of groups formed for the three companies will be 5, 8, and 6, respectively. Therefore, the least number of total groups formed = 5 + 8 + 6 = 19.

**Question 2:**

The numbers 2604, 1020 and 4812 when divided by a number N give the same remainder of 12. Find the highest such number N.

**Answer & Solution**

Since all the numbers give a remainder of 12 when divided by N, hence (2604 – 12), (1020 – 12) and (4812 – 12) are all divisible by N. Hence, N is the HCF of 2592, 1008 and 4800. Now 2592 = 2^{5}× 3^{4}, 1008 = 2^{4}× 3^{2}× 7 and 4800 = 2^{6}× 3 × 5^{2}. Hence, the number N = HCF = 2^{4}× 3 = 48.

**Question 3:**

The numbers 400, 536 and 645, when divided by a number N, give the remainders of 22, 23 and 24 respectively. Find the greatest such number N.

**Answer & Solution**

N will be the HCF of (400 – 22), (536 – 23) and (645 – 24). Hence, N will be the HCF of 378, 513 and 621. \(\Rightarrow N = 27\)

**Question 4:**

The HCF of two numbers is 12 and their sum is 288. How many pairs of such numbers are possible?

**Answer & Solution**

If the HCF if 12, the numbers can be written as 12x and 12y, where x and y are co-prime to each other. Therefore, 12x + 12y = 288 \(\Rightarrow x + y = 24\)

The pair of numbers that are co-prime to each other and sum up to 24 are (1, 23), (5, 19), (7, 17) and (11, 13). Hence, only four pairs of such numbers are possible. The numbers are (12, 276), (60, 228), (84, 204) and (132, 156).

**Question 5:**

The HCF of two numbers is 12 and their product is 31104. How many such numbers are possible?

**Answer & Solution**

Let the numbers be 12x and 12y, where x and y are co-prime to each other. Therefore, 12x × 12y = 31104
\(\Rightarrow xy = 216\).

Now we need to find co-prime pairs whose product is 216.

216 = 2^{3}× 3^{3}. Therefore, the co-prime pairs will be (1, 216) and (8, 27). Therefore, only two such numbers are possible.

**Question 6:**

Find the HCF of 2

^{100}– 1 and 2

^{120}– 1

[1] ${{2}^{20}}-1$

[2] ${{2}^{10}}-1$

[3] ${{2}^{40}}-1$

[4] 1

**Answer & Solution**

2^{100}– 1 = (2^{20})^{5}- 1 $\Rightarrow$ divisible by 2^{20 }- 1 (a^{n}- b^{n} is always divisible by a - b)

Similarly, 2^{120}– 1 = (2^{20})^{6}- 1 $\Rightarrow$ divisible by 2^{20 }- 1 (a^{n}- b^{n} is always divisible by a - b)

$\Rightarrow$ HCF = 2^{20 }- 1

**Question 7:**

Find the highest four-digit number that is divisible by each of the numbers 24, 36, 45 and 60.

**Answer & Solution**

24 = 2^{3}× 3, 36 = 2^{2}× 3^{2}, 45 = 3^{2}× 5 and 60 = 2^{3}× 3^{2}× 5. Hence, the LCM of 24, 36, 45 and 60 = 2^{3}× 3^{2}× 5 = 360. The highest four-digit number is 9999. 9999 when divided by 360 gives the Remainder 279. Hence, the number (9999 – 279 = 9720) will be divisible by 360. Hence the highest four-digit number divisible by 24, 36, 45 and 60 = 9720.

**Question 8:**

Find the highest number less than 1800 that is divisible by each of the numbers 2, 3, 4, 5, 6 and 7.

**Answer & Solution**

The LCM of 2, 3, 4, 5, 6 and 7 is 420. Hence 420, and every multiple of 420, is divisible by each of these numbers. Hence, the number 420, 840, 1260, and 1680 are all divisible by each of these numbers. We can see that 1680 is the highest number less than 1800 which is multiple of 420. Hence, the highest number divisible by each one of 2, 3, 4, 5, 6 and 7, and less than 1800 is 1680.

**Question 9:**

Find the lowest number which gives a remainder of 5 when divided by any of the numbers 6, 7, and 8.

**Answer & Solution**

The LCM of 6, 7 and 8 is 168. Hence, 168 is divisible by 6, 7 and 8. Therefore, 168 + 5 = 173 will give a remainder of 5 when divided by these numbers.

**Question 10:**

What is the smallest number which when divided by 9, 18, 24 leaves a remainder of 5, 14 and 20 respectively?

**Answer & Solution**

The common difference between the divisor and the remainder is 4 (9 - 5 = 4, 18 - 14 = 4, 24 - 20 = 4). Now the LCM of 9, 18, and 24 is 72. Now 72 - 4 = 72 - 9 + 5 = 72 - 18 + 14 = 72 - 24 + 20. Therefore, if we subtract 4 from 72, the resulting number will give remainders of 5, 14, and 20 with 9, 18, and 24.

Hence, the number = 72 - 4 = 68.

**Question 11:**

A number when divided by 3, 4, 5, and 6 always leaves a remainder of 2, but leaves no remainder when divided by 7. What is the lowest such number possible?

**Answer & Solution**

The LCM of 3, 4, 5 and 6 is 60. Therefore, the number is of the form 60k + 2, i.e. 62, 122, 182, 242 etc. We can see that 182 is divisible by 7. Therefore, the lowest such number possible = 182.

**Question 12:**

For how many ordered pairs (

*a*,

*b*) of natural numbers is the LCM of

*a*and

*b*is 2

^{3}5

^{7}11

^{13}?

**Answer & Solution**

Let's solve for the powers of 2. One of the number will have 2^{3} in it, as the LCM has 2^{3}.

Now the other number can have the powers of 2 as 2^{0}, 2^{1}, 2^{2}, and 2^{3}. Therefore, number of pairs will be 4: (2^{3}, 2^{0}), (2^{3}, 2^{1}), (2^{3}, 2^{2}), and (2^{3}, 2^{3}) and the number of ordered pairs will be 2 × 4 - 1 = 7 (we cannot count the pair (2^{3}, 2^{3}) twice.

Similarly ordered pairs for powers of 5 = 2 × 8 - 1 = 15.

Number of ordered pairs for powers of 7 = 2 × 14 - 1 = 27.

Total ordered pairs (a, b) = 7 × 15 × 27 = 2835