Competitive exams often ask questions with a 'wrapper' around them. Its important to get to the right question quickly. Give the underlying questions for the following -
1. Give the smallest 4-digit number with an odd number of factors (easy one)
2. Set S contains elements { 6300, 2100, 1260, 900, 700, 6300/11, 6300/13, 420, ..........}, how many elements of Set S are integers?
3. f(k) gives the sum of all digits of number k. g(k) = f(f(f(...k))) such that we end up with a number from 1 to 9. How many 5-digit numbers n exist such that g(n) = 2
Discussion:
The questions sitting underneath the above statements are as follows
1. Give the smallest 4-digit number with an odd number of factors (easy one) => What is the smallest 4-digit perfect square?
2.
Set S contains elements { 6300, 2100, 1260, 900, 700, 6300/11, 6300/13,
420, ..........}, how many elements of Set S are integers? => How many odd factors does 6300 have?
3.
f(k) gives the sum of all digits of number k. g(k) = f(f(f(...k))) such
that we end up with a number from 1 to 9. How many 5-digit numbers n
exist such that g(n) = 2? => How many 5-digit number exist that when divided by 9 leave a remainder of 2?
Explanatory Answers
1. Give the smallest 4-digit number with an odd number of factors (easy one) => What is the smallest 4-digit perfect square? 1024
2.
Set S contains elements { 6300, 2100, 1260, 900, 700, 6300/11, 6300/13,
420, ..........}, how many elements of Set S are integers? => How many odd factors does 6300 have?
6300 = 2^2 * 3^2 * 5^2 * 7. Number of odd factors of this number = (2+1) * (2+1) * (1+1) = 18. For discussion on number of odd factors, look
here .
3.
f(k) gives the sum of all digits of number k. g(k) = f(f(f(...k))) such
that we end up with a number from 1 to 9. How many 5-digit numbers n
exist such that g(n) = 2? => How many 5-digit number exist that when divided by 9 leave a remainder of 2?
There are 90000 5-digit numbers. There will be 10000 numbers that leave a remainder of 2 on division by 9 within these. Answer = 10000.
g(n) mentioned above is identical to the remainder when a number is divided by 9. Once you pick that, this question become a sitter. Competitive exams are good at masking questions. As much as possible, learn from first principles. If one had thought about why the test of divisibility for 9 works, this bit would have been clear.
CAT
2013 Course offered by 2IIM @ ChennaiCourse handled by IIM Alumni, Batches @
Anna Nagar, Mylapore and Velachery.
Weekend
batches @ Anna Nagar from 24th November, @ Mylapore from 17th November, @
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