Something about numbers

Milind requested something about numbers and probability. Since probability is not too crucial as far as CAT goes, I will tackle numbers.

Number Systems is the topic with highest number of questions in CAT. Right since 1999, there have been around 11 to 13 Qs out of the 50 Qs. Even in the last couple of years when the Qs decreased to 30, number systems still had about 7-8 questions.

Also, I see most of the aspirants stuck with this topic even several months after they have started their studies. Even today when i take revision sessions, students request to start with numbers. This must tell you enough about the width and depth of the topic. It is one topic that gives one an immense amount of confidence if one is good at. So do write in about any type of problem or any specific problem for me to take on from.

Right now will just explain a good link between basic fundas and very tough looking questions and their explanations. Will build it through examples.

Find the general form of numbers which when divided by 8 leave a remainder of 3 and when divided by 11 leave a remainder of 10. Work on this example before reading ahead.

Rather than any maths, i would suggest start with the larger divisor (why? think about this after reading ahead)

Find numbers that are divisible by 11. Obviously 11, 22, 33, 44, 55, ...
On dividing by 11, there sould be a remainder of 10. Simply add 10 to the above numbers. This will remain as remainder. Obviously such numbers are 21, 32, 43, 54, ....

Take each number turn by turn, and check in your mind which satisfies the other condition. Obviously the smallest number satisfing both the conditions is 43. Now lets now find the next and the series of numbers satisfying the two conditions.

43 when divided by 11 leaves remainder 10. What should be added to 43, so that the condition stays? Obviously 11 or 22 or 33 i.e. multiple of 11.

Similarly, 43 when divided by 8 leaves remainder 3. Only when we add 8 or a multiple of 8 to 43, the number so obtained would also leave remainder of 3.

For both the conditions to be valid, a multiple of 8 and 11 should be added to 43. Thus general form of the number is 88n + 43.

Now on to a seemingly difficult question.
Find the remainder when 12345678910111213.....9899100101 (first 101 natural numbers written side by side) when divided by 36.

Let N be 123456789101112.....9899100101.
N when divided by 4 leaves a remainder of 1 and when divided by 9 leaves a remainder of 3 (how can this be identified immediately? read points to ponder towards end of the post if not clear immediately).

Find the general form of numbers which when divided by 4 leave remainder 1 and when divided by 9 leaves a remainder of 3.

Now, the general form of such numbers, as per the above explanation, should easily be found to be 36n + 21. (One should just check 3, 12, 21 turn by turn to identify the first number as 21).

Thus N is of the form 36n + 21. So when divided by 36 leaves a remainder of 21.

Points to ponder and learn:
1. Why was the divisor 36 factorised as 4 and 9? Could we use 3 and 12 instead of 4 and 9?
2. Is the sum of digits of 1234567891011....9899100101 same as E101 = (101 x 102)/2?
3. If not, how can we immediately find the remainder when N is divided by 9? Clue: In any set of 100 consecutive numbers, how many times does any digit (say 5 or 7) appear in units and tens place?
4.
Numbers that leave a remainder of 10 when divided by 11 are 10, 21, 32, 43, ... This is a AP with c.d. 11
Numbers that leave a remainder of 3 when divided by 8 are 3, 11, 19, 27, ... This is an AP with c.d. 8.
We just found numbers that are common to both the APs. And such numbers themselves form a AP with c.d. LCM of 8 and 11 i.e. 88
Hope you see how the above question can be asked in an altogether different format and topic!
5. Writing the above data in algebra, we have 11a + 10 = 8b + 3, where a and b are whole numbers. Re-arranging we have 8b - 11a = 7. Find whole number solutions to this equation. What pattern does a and b follow? Next posting will be on this aspect of integer solution to a equation in two variables.
6. Knowing that numbers leaving a remainder of 10 and 3 when divided by 11 and 8 respectively is 88n + m (let's say we do not spend time in finding m), find how many such three digit numbers exist? Can this question be done immediately, without any work? (Please use the fact that there are 900 three digit numbers)
7. Why to start with the larger divisor?

Happy pondering!

Chandra

If you know how to find remainders when a^b (a raised to b) is divided by some number, try the following to consolidate the above learning.
Find the remainder when 2^500 is divided by 153.