# IIM CAT Preparation Tips

IIM CAT Preparation Tips: March 2011

## Mar 30, 2011

### CAT Number Theory - Challenging question

Just one question this time, but a fairly challenging one though.

There is a 4-digit number 'abcd' that satisfies the following property. 'abcd' = ab ^2 + cd ^2. Find abcd.

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### Answers to World Cup questions

Have given below the answers to the world cup questions. The questions can be found here .

1. Australia, in one of their matches, scored a total that is the lowest multiple of 11 with exactly 10 factors. Against who was this?

2^4 * 11 = 176. Australia scored this against Pakistan

2. In one of the matches, the top-scorer (who scored a century) for the team batting first scored 3/8th the team's runs, but the team still did not cross 300. The HCF of the team's score and the top-scorers score is a prime number. Which match was this?

Top-scorer's score is a multiple of 3 greater than 100. So, it has to be a score like 34 * 3, 35 * 3, 36 * 3,..... The overall score is less than 300. So, the overall score has to be lower than 38 * 8. So, the overall score has to be of the form 37 * 8, 36 * 8, 35 * 8,...

The HCF is a prime number, this implies that the top scorer scored 111, the team scored 296. India vs. South Africa. Sachin Tendulkar century

3. A 'minnow' scored a total that can be made a perfect square if it is multiplied by 23. If they had scored 1 more, then their score multiplied by 13 would have yielded a perfect square. What was their score? Bonus point, what match was this?

The score should have been 23 * 4 or 23 * 9, 23 * 9 = 207. 207 + 1 = 208, which is 13 * 16.
India vs. Ireland. Ireland scored 207

4. In a particular match, team A batting first scored twice a perfect square. Team B batting second, also scored twice a perfect square. The total score of both the teams put together was also a perfect square. What match are we talking about here?

India 338, England 338 (169 * 2). Total 676 (26* 26)

5. Team scored a three digit score 'abc' in a match. a and b are prime numbers. a + 2b and a + 2c are also prime. b + 2a and b + 2c are also primes. a + c and b + c are also primes. Two digit numbers 'ab' and 'ba' are both primes. Which match was this?

a + c and b + c are also primes => a, b should be odd, c should be even
a,b cannot be 5 as ab is also prime. So, abh has to be 37.
c can only be zero.
abc = 370 (India vs. Bangladesh)

## Mar 26, 2011

### World Cup Special - Few maths questions

My colleague mentioned that he had asked his class 1-2 questions on numbers related to the WC. Just thought it might be a good idea to follow that.

1. Australia, in one of their matches, scored a total that is the lowest multiple of 11 with exactly 10 factors. Against who was this?

2. In one of the matches, the top-scorer (who scored a century) for the team batting first scored 3/8th the team's runs, but the team still did not cross 300. The HCF of the team's score and the top-scorers score is a prime number. Which match was this?

3. A 'minnow' scored a total that can be made a perfect square if it is multiplied by 23. If they had scored 1 more, then their score multiplied by 13 would have yielded a perfect square. What was their score? Bonus point, what match was this?

4. In a particular match, team A batting first scored twice a perfect square. Team B batting second, also scored twice a perfect square. The total score of both the teams put together was also a perfect square. What match are we talking about here?

5. Team scored a three digit score 'abc' in a match. a and b are prime numbers. a + 2b and a + 2c are also prime. b + 2a and b + 2c are also primes. a + c and b + c are also primes. Two digit numbers 'ab' and 'ba' are both primes. Which match was this?

## Mar 16, 2011

### Solutions to questions from assorted topics

Have given below solutions to the questions given here .

1. A number n! is written in base 6 and base 8 notation. Its base 6 representation ends with 10 zeroes. Its base 8 representation ends with 7 zeroes. Find the smallest n that satisfies these conditions. Also find the number of values of n that will satisfy these conditions.

Base 6 representation ends with 10 zeroes, or the number is a multiple of 6^10. If n! has to be a multiple of 6^10, it has to be a multiple of 3^10. The smallest factorial that is a multiple of 3^10 is 24!. So, when n = 24, 25 or 26, n! will be a multiple of 6^10 (but not 6^11).

Similarly, for the second part, we need to find n! such that it is a multiple of 2 ^ 21, but not 2 ^ 24. When n = 24, n! is a multiple of 2^22. S0, when n = 24, 25, 26, 27, n! will be a multiple of 2 ^ 21 but not 2 ^ 24.

The smallest n that satisfies the above conditions is 24. n = 24, 25 or 26 will satisfy the above conditions.

2. [x] is the greatest integer less than or equal to x. Find the number of positive integers n such that [n/11] = [n/13].

This is a classic case of brute-force counting.
When n = 1, 2, 3....10, both values will be equal to 0. 10 possibilities
When n = 13,14, ....21, both values will be equal to 1. 9 possibilities
When n = 26,27, ....32, both values will be equal to 2. 7 possibilities
When n = 39,27, ....43, both values will be equal to 3. 5 possibilities
When n = 52, 53, 54, both values will be equal to 4. 3 possibilities
when n = 65, both will be equal to 5. 1 possibility

So, totally there are 1 + 3 + 5 + 7 + 9 +10 = 35 possibilities.

3. Positive numbers 1 to 55, inclusive are placed in 5 groups of 11 numbers each. What is the maximum possible average of the medians of the 5 groups?

We need to maximise each median in order to have the overall maximum median possible.

The highest possible median is 50 as they should be 5 numbers higher than the median in the group of 11. So, if we have a set that has a, b, c, d, e, 50, 51, 52, 53, 54, 55, the median will be 50. In this set, it is best not to waste any high values on a, b, c, d or e as these do not affect the median. So, a set that reads as 1, 2, 3, 4, 5, 50, 51, 52, 53, 54, 55 will also have median 50.

The next set can be 6, 7, 8, 9, 10, 44, 45, 46, 47, 48, 49. The median will be 44.

The medians of the 5 groups can be 50, 44, 38, 32, 26. The highest possible average of the medians will be 38.

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## Mar 8, 2011

### Few more questions - assorted topics

Have given below a few more questions from assorted topics

1. A number n! is written in base 6 and base 8 notation. Its base 6 representation ends with 10 zeroes. Its base 8 representation ends with 7 zeroes. Find the smallest n that satisfies these conditions. Also find the number of values of n that will satisfy these conditions.

2. [x] is the greatest integer less than or equal to x. Find the number of positive integers n such that [n/11] = [n/13].

3. Positive numbers 1 to 55, inclusive are placed in 5 groups of 11 numbers each. What is the maximum possible average of the medians of the 5 groups?

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## Mar 5, 2011

### Solutions - Mixed bag

Have given solutions to the questions posted here

1. Average of 6 distinct positive integers is 33. The median of the three largest numbers is 43. What is the difference between the highest and lowest possible median of the 6 numbers?

Let a, b, c, d, e, f be the six numbers in ascending order

e = 43; a + b + c + d + e + f = 198

Median of 6 integers = (c +d)/2

(c +d)/2 is maximum when c and d are maximum since e = 43, dmax = 42 and cmax = 41

(c +d)/2 is minimum when c and d are minimum amin = 1, bmin = 2, cmin = 3 dmin = 4

Therefore, = (c +d)/2 min = 3.5

Max – min = 41.5 – 3.5 = 38

2. In class A, the ratio of boys to girls is 2:3. In class B the ratio of boys to girls is 4 : 5. If the ratio of boys to girls in both classes put together is 3 : 4, what is the ratio of number of girls in class A to number of girls in class B?

Let us assume class A has 2x boys and 3x girls, class B has 4y boys and 5y girls.

(2x + 4y)/(3x + 5y) = ¾

8x + 16y = 9x + 15y

x = y

Reqd ratio = 3x/5y (since x = y) = 3/5

3. N is an 80-digit positive integer (in the decimal scale). All digits except the 44th digit (from the left) are 2. If N is divisible by 13, find the 26th digit.

To begin with, the question should read "find the 44th digit".

Any number of the form abcabc is a multiple of 1001. 1001 is 7 * 11 * 13. So, any number of the form abcabc is a multiple of 13.

So, a number comprising 42 2's would be a multiple of 13, so would a number comprising 36 2's. So, in effect, we are left with a two digit number 2a, where a is the 44th digit. 26 is a multiple of 13, so the 44th digit should be 6.

4. A page is torn from a novel. The sum of the remaining digits is 10000. What is the sum of the two page-numbers on the torn page of this novel?

n(n+1)/2 should be nearby 10,000, so n(n+1) is somewhere near 20,000. So n should be around sqrt(20,000) about 141. Try 142

142*143 = 10153

141 *142 = 10011

So, the missing pages are either 76 and 77 or 5 & 6

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