Solved CAT Questions (Number Theory)  Set 1

Q1. Which among 2^(1/2), 3^(1/3), 4^(1/4), 6^(1/6), 12^(1/12) is the largest?
a. 2 1/2
b. 3 1/3
c. 4 1/4
d. 6 1/6
e. 12 1/12 (CAT 2006)Solution: The common approach is to equate the exponents and then compare. But will learn a much faster/easier trick to solve such questions first and then will go the the traditional method.
Given numbers are in the form x^1/x[Concept: f(x) = x^1/x is maximum at x = e ( e = 2.71) and f(x) will decrease if you go either side of x = e]
So f(x) has a higher value if x is nearer to e ( check whether all the given numbers are either greater than or lesser than e )
Here in our case x = 2, 3, 4, 6 and 12. except 2 all other numbers are greater than e so our only job here is to compare 2^1/2 and 3^1/3 as we know that 3^1/3 > 4^1/4 > 6^1/6 > 12^1/2 ( because e < 3 < 4 < 6 < 12 )2^1/2 : 3^1/3 = 2^3 : 3^2 = 8 : 9 (equate the exponents by raising both term to the power of 6, LCM ( 2 ,3) )
so 3^1/3 is the largest.PS: we can use the same trick to compare a^b and b^a
Closer the base (a or b) is to e, higher the value will be.
example:
9.1 ^ 8.9 and 8.9 ^ 9.1
both these values are greater than e, and 8.9 is closer to e than 9.1, so 8.9^9.1 is higher than 9.1^8.9another classic one,
which is greater, e^PI or PI^e ?
we know e = 2.7 and PI = 3.14 => e^PI is greater.
This trick is especially handy when all the given numbers belongs to the same side of e, where we can do the comparison in no time.Now traditional approach  as promised:
We are going to equate the exponents (powers), and compare the bases. Here the given exponents are 1/2, 1/3, 1/4, 1/6 and 1/12. LCM of 2, 3, 4, 6, and 12 is 12.
Raise every number to power of 12 => it becomes 2^6 , 3^4 , 4^3 ,6^2 ,12 => 64, 81, 64, 36, 12
So, the biggest is 81^1/12, and hence 3^1/3.Also, here, if you observe carefully, 2^1/2 and 4^1/4 are the same and since the options have both of them, they can be eliminated from calculation, and you need to calculate LCM and bases for only 3 options.
Q2. If x = − 0.5, then which of the following has the smallest value?
a. 2^(1/x)
b. 1/x
c. 1/(x^2)
d. 2^x
e. 1/√x (CAT 2006)Solution: We need to calculate the smallest value. Before starting the calculation right away, spend at least 10 seconds on analysis, to see if you can solve the question without big calculations.
We know that x =  0.5
If we have x as the exponent, we know that even if x is negative or positive, the answer is positive. i.e. If x is positive, 2^x is positive.
If x is negative (let it be represented by –x), 2^(x) is same as 1/2^x and hence it is positive.
1/x^2 is positive. Even numbers as exponents always results in positive value.
Option e is also positive, as (x) is involved and it turns out to be positive.
Option b is the only negative number involved in the options, and hence the answer.Q3. If a/b = 1/3, b/c = 2, c/d = 1/2, d/e = 3 and e/f = 1/4 then what is the value of abc/def
(1) 3/8
(2) 27/8
(3) 3/4
(4) 27/4
(5) 1/4 (CAT 2006)Solution:
Method 1: We can use ratios
a : b : c : d : e : f = 1 : 3 : 3/2 : 3 : 1 : 4
So abc/def = (1 * 3 * 3/2)/(3 * 1 * 4) = 3/8Method2: abc/def is same as a/d * b/e * c/f
a/d = a/b * b/c * c/d = 1/3 * 2 * 1/2 = 1/3
b/e = b/c * c/d * d/e = 2 * 1/2 * 3 = 3.
c/f = c/d * d/e * e/f = 1/2 * 3 * 1/4 = 3/8.
abc/def = a/d * b/e * c/f = 1/3 * 3 * 3/8 = 3/8.Q4. Consider a sequence, where the nth term is tn = n/ (n+2) , n = 1, 2….
The value of t3 * t4 * t5 * … t53 is
(1) 2/495
(2) 2/477
(3) 12/55
(4) 1/1485
(5) 1/2970 (CAT 2006)Solution:
tn = n/n+2. If n = 1, t1 = 1/3. Similarly t2 = 2/4, t3 = 3/5 and so on.
t3 = 3/5, t4 = 4/6, t5 = 5/7…….
We need to find t3 * t4* t5*……..t53
i.e. 3/5 * 4/6 * 5/7 * 6/8 *………………..51/53 * 52/54 * 53/55
i.e. 3/5* 4/6*5/7*6/8*………………..51/53*52/54 *53/55We see that except 3 and 4 in the numerator everything else cancels off, with the previous denominators. So, we have only 3 * 4 in the numerator. Similarly, in the denominators, except for the last 54 and 55, everything else cancels off. So only 54 * 55 in the denominator. So, the answer will be 3 * 4 / 54 * 55. 54 is divisible by 6. So 6 can be cancelled from numerator and denominator. The remaining when multiplied gives 2 / 9 * 55 = 2/495.
Q5. The sum of four consecutive two digit odd numbers, when divided by 10, becomes a perfect square. Which of the following can possibly be one of these four numbers?
(1) 21
(2) 25
(3) 41
(4) 67 (CAT 2006)Solution: One can observe that there are 2 ways to go about it. Whichever way it is, you have to get a start, and that is : 4 consecutive odd numbers can end with 1 among :
1, 3, 5, 7 – Sum of unit’s place not divisible by 10
3, 5, 7, 9 – Sum of unit’s place not divisible by 10
5, 7, 9, 1 – Sum of unit’s place not divisible by 10 (here the last 1 is the one in the next cycle. e.g. 15, 17, 19, 21)
7, 9, 1, 3 – Sum of unit’s place is divisible by 10 (here the last 1 & 3 is the one in the next cycle. e.g. 17, 19, 21, 23)
9, 1, 3, 5 – Sum of unit’s place not divisible by 10 ( last 1, 3, 5 like above 2 cases, next cycle, eg:19, 21, 23, 25)Since, we know that the number when divided by 10 gives a perfect square, the unit’s place sum should add up to a multiple of 10. The only possible combination of unit digit of numbers is (7, 9, 1, 3)
First approach : Take the options 1 by 1.
1st option : 21 : So the numbers are 17, 19, 21, 23. Take sum, divide by 10 and see if the quotient is a perfect square.
2nd option : 25 : No need to take this option at all, as the numbers involved will definitely end in 7, 9, 1, 3 and 25’s unit place is 5.
3rd option : 41 : Numbers are 37 , 39, 41, 43. Find sum, divide by 10, and see if quotient is perfect square.
4th option : 67 : Numbers are 67, 69, 71, 73. Find sum, divide by 10, and see if quotient is perfect square.Second approach : The numbers end in 7, 9, 1, 3. Let the ten’s place be x. So, the numbers are 10x + 7, 10x + 9, 10x + 11 and 10x + 13 (remember it is not 10x + 1 and 10x + 3, as the consecutive odd digits after 9 is 11, and then 13)
Their sum is 40x + 40.
Quotient when divided with 10 is 4x + 4 or 4 * (x+1).
4 * (x+1) is a perfect square.
x can be 3 or 8.
If x is 3, numbers are 37, 39, 41, 43
If x is 8, numbers are 87, 89, 91, 93Q6. When you reverse the digits of the number 13, the number increases by 18. How many other two digit numbers increase by 18 when their digits are reversed?
(1) 5
(2) 6
(3) 7
(4) 8
(5) 10 (CAT 2006)Let the unit’s place of the original number be y and ten’s place be x.
So, the number is 10x + y.
After reversing, the unit’s place will be x and ten’s place y. Number will be 10y + x.
We need to know, how many numbers when reversed will give an increase of 18.
So, 10y + x – (10x + y) = 18 ; 9 (y – x) = 18 ; y  x = 2
y cannot be 2, as x will be 0 and the number will be 02, which is not a 2 digit number.
y can take values from 3 – 9, and x can take values from 1 – 7 (y = 3, x = 1……y = 4, x = 2 and so on…)
So, there are 6 numbers other than 13 which will satisfy the condition.Q7. The number of employees in Obelix Menhir Co. is a prime number and is less than 300. The ratio of the number of employees who are graduates and above, to that of employees who are not, can possibly be:
(1) 101 : 88
(2) 87 : 100
(3) 110 : 111
(4) 85 : 98
(5) 97 : 84 (CAT 2006)Solution: The total number of employees will be the sum of graduates and nongraduates and this should be a prime number, and less than 300. To find if the number is prime or not, these are the simple divisibility tests which I use.
2 (if it is even, it is divisible by 2, else not, so ending with 0, 2, 4, 6, 8 divisible by 2)
3 (If sum of digits divisible by 3, then number is divisible by 3)
5 (Number will end with 5 or 0)
7 ( I will rather divide and see)
11 (Add alternate digits, and subtract with the next set of alternate digits. If the difference is 0/multiple of 11, then divisible by 11)
Also, you need to see if the numbers are divisible till their square root (approximately, at least). After their square root, they will not be divisible by any more numbers.
For eg : Consider 70 : Square root of 70 is between 8 & 9. So, we need to check till 8 if all the numbers are divisible or not. (i.e.2, 3, 5, 7). If none of the prime numbers till the square root are not divisible, then the number is a prime number.Consider the options:
101 : 88, sum is 189. 189 is divisible by 3 (Sum of digits divisible by 3)
87 : 100, sum is 187, 187 is divisible by 11
110 : 111, sum is 221. 221 is divisible by 13. (check prime numbers till 15, as 15^2 is 225 > 223)
85 : 98, sum is 183, divisible by 3.
97 : 84, sum is 181, is a prime number. (Check all prime numbers till 14, as 14^2 is 196 > 181)Note: Any prime no: > 3 can be written in the form 6k + 1 or 6k  1. So, out of the given options check if we would be able to write the sum in that format. Option 2 and 5 satisfies and now we can just check only these 2 options for prime. Option 2 is divisible by 11 so answer should be option 5.
Q8. Consider four digit numbers for which the first two digits are equal and the last two digits are also equal. How many such numbers are perfect squares?
(1) 3
(2) 2
(3) 4
(4) 0
(5) 1 (CAT 2007)Solution:
Method1: We know that only for 12^2 last 2 digits are equal.use this property.since number is a 4 digit number the square root has to be a 2 digit number. We know (5012)^2, (50+120^2, (10012)^2, (100+12)^2 have same last 2 digits as 12^2.but last one will yield a 5 digit number. Just exclude it and check the rest 3. You'll get only (10012)^2 = 88^2 satisfying the criteria. so option 5
Method 2: Let the number be aabb.
aabb = 1000a + 100a + 10b + b = 1100a + 110 b.
1100 and 110 are divisible by 11 and 11 is a prime number. i.e. 1100a + 11b = 11 (100a + b)
Since aabb sould be a perfect square, and 11 is already a factor of that, there should be “one more” 11 that should be a factor. (so that it becomes a perfect square)
So, 121 should be a factor of aabb.
The number can be 121 * x, where x should be another perfect square.
The numbers could be : 121 * 9, 121 * 16, 121 * 25, 121 * 36, 121 * 49, 121 * 64, 121 * 81
Out of this, 121 * 64 is 7744, which is of the form aabb. None of the others are.Q9. All the page numbers from a book are added, beginning at page 1. However, one page number was mistakenly added twice. The sum obtained was 1000. Which page number was added twice?
(1) 44
(2) 45
(3) 10
(4) 12 (CAT 2001)Solution: Sum of n natural numbers = n (n+1)/2
here, n( n + 1)/2 = 1000  x (where x is the number that is added twice)
n(n+1) < 2000  2x
we need to find max value n can take . We know for a product of two numbers to be max the number should be closest... find out the least number whose square is higher than 2000.
45^2 = 2025Will break here to learn a neat trick
45^2 = [ 4 (digit(s) excluding 5) * (4 (digit(s) excluding 5) + 1)] [25] = 2025.
Similarly 135^2 = [13 x 14] [25] = 18225. Got it ? Good, back to our problem!So our number is somewhere near to 45.. it cant be more than 45 as our required product is less than 2000.
44 * 45 = 1980.. Cool! so we can have n = 44
n (n + 1) = 44 * 45 = 1980 = 2000  2 * 10
x = 10 ( so we have 44 pages and page number 10 was added twice) Option 3Q10. For a Fibonacci sequence, from the third term onwards, each term in the sequence is the sum of the previous two terms in that sequence. If the difference in squares of seventh and sixth terms of this sequence is 517, what is the tenth term of this sequence?
(1) 147
(2) 76
(3) 123
(4) Cannot be determined ( CAT 2001 )f(7)^2  f(6)^2 = 517
(f(7) + f(6)) (f(7)  f(6)) = 517 = 11 x 47
As 11 and 47 cannot be further factorized and f(7) or f(6) should be natural numbers, we can say
f(7) + f(6) = 47 (Largest prime factor should come from their sum)
f(7)  f(6) = 11 (smallest prime factor should come from their difference)
solving, f(7) = 29 and f(6) = 18f(10) = f(8) + f(9)
= f(7) + f(6) + f(7) + f(8)
= f(7) + f(6) + f(7) + f(7) + f(6)
3f(7) + 2f(6) = 3 x 29 + 2 x 18 = 123.