GRE Plus Fun Mathematics

Determining the remainder...Next

2006-03-28T22:52:00.000-08:00

Yesterday we had two problems:
P1. What is remainder when 8^102 is divided by 7?
P2. What is remainder when 9^55 is divided by 4?

Answers are:
P1. First discover the cycle for powers of 8 as divided by 7.
So 8^1 divided by 7 gives remainder 1.
This implies all powers of 8 shall yield remainder when divided by 7.
So answer is 1.

P2. 9^1 by 4 gives remainder 1.
This is similar to P1.
So answer is 1.

Let us have few more problems:

P3. What is the remainder when 6^127 is divided by 13?
P4. What remainder is generated when you divide 5^125 by 124?
P5. Can you find out what remainder is there in the division of 3^1000 by 7?

Answers and Solution in next posting.

Squaring Numbers Close to 50

2005-10-19T03:28:00.000-07:00

Take 47. This is deficit 3 from 50.
Less 3 from 25 to get 22 and to its right place 3 square i.e. 09.
So we get 2209.

For 41, deficit is 9.
So 25-9 = 16 and place 81 to its right to get 1681.

For 57, add 7 to 25 to obtain 32 and place 47 to its right:
3249.

In case of 61, add 11 to 25 to have 36. 11 square is 121, use just two digits 21 and carry the extra 1 to 36 to make it 37. So square is 3721.

Sqauring Numbers Cloase to 1000

2005-10-18T03:33:00.000-07:00

Take 978. It is at a deficit of 22 from 1000.
Square it to and you get 484, place these as right three digits of the answer (three digit because 1000 has three zeroes).
On left, reduce 978 by 22 to obtain 956. So the correct response is 956484.

For 991, deficit is 9 and its square is 81, which we use as 081 as we want three digits.
Left is 991-9 = 982. So square is 982484.

For 1022, add 22 to 1022 to have 1044 and to its right attach 484. The answer is 1022484.
For 1031, we have 1062961 as the square since 31 is added to 1031, and 31 square is 981.

Squaring: Keeping A Mirror At 25

2005-10-17T21:16:00.000-07:00

Let me write a few squares around the number 25.

20^2 = 400
21^2 = 441
22^2 = 484
23^2 = 529
24^2 = 576
25^2 = 625
26^2 = 676
27^2 = 729
28^2 = 784
29^2 = 841
30^2 = 900

Can you see a similarity between numbers equally spaced on either side of 25.

So if you somehow memorize squares of numbers 1 to 25,
you will find that higher numbers can be squared very easily.

e.g. For 29, which is 4 above 25, go 4 below 25, i.e. to 21. 21^2 is 441, to this add 400 to get 841.

For 27, which is 2 above 25, go 2 below 25 to 23. Now 23 square is 529 to which you add 200 to get 729.

In fact if you want to know square of, say 39, which is 14 above 25, go 14 below 25, i.e. to 11. Add 1400 to 11 square (=121) to get 1521.

To square 49, which is 24 above 25, add 2400 to the square of number 24 below 25, which is 1. So 1^2 is 1 and adding 2400, we get 2401.

Ain't this great. As if there is a mirror at 25.

In fact if you want square of 59, which is 34 above 25, we add 3400 to square of number 34 below 25, which is -9. But square of -9 is 81, add this to 3400 to yield 3481.

Trying again for 73, we have 73 exceeding 25 by 48.
Now 4800 has to be added to what? Less 48 from 25 to get -23. Square of 23 we saw above is 529. We add this to 4800 to get 5329.

The Algebra Behind Squaring Numbers Near 100

2005-10-16T00:27:00.000-07:00

The previous two posts may have seemed magical to many.

This however is plain algebra, the (a+b)^2 formula.

Let a number less than 100 be called (100 - d)
e.g. 93 is 100-7.

So squaring this, we get 100000 - 200d + d^2 or 100(100 - d - d) + d^2.
Isn't this what we have been doing, taking the square of d (7 in case of 93) and placing its square d^2 (or 49 in the example) to the right; since the remaining part of the answer is multiplied by 100, it leaves the right two digits untouched anyways.

And within the brackets, we have been reducing the number (=100 - d) by a further d.

In a similar vein, a number exceeding 100 would be written as (100 + e).

Its square is 10000 + 200e + e^2 or 100(100 + e + e) + e^2.
This you would see is what we did for numbers exceeding 100.

Squaring 100+ Numbers

2005-10-15T00:02:00.000-07:00

Let us say you want to square 104.
This is 4 in excess of 100, square 4 to get 16, this gives you the right part of the answer.
Just add this 4 to 104 to obtain 108, which is to placed to left of 16 you got above,
voila! 10816 is the answer.

For 109, 9 is the excess on 100, so right part is 81. Left part is 9 added to 9, i.e. 118.
So the answer is 11881.

For 112, 12 is the excess or surplus on 100. Its square is 144, of which 44 form the right two digits of the answer, and 1 is carried to the left.
Left becomes 12 + 112, i.e. 124, with the carried 1 added, it is 125.
So the answer is 12544.

Similarly 103 square is found by squaring the surplus 3, i.e. 9 which we write as 09 as two digits have to be occupied.
The left is 3 + 103 = 106.
So the correct answer is 10609.

Squaring Carries On

2005-10-14T15:55:00.000-07:00

Take a number like 94, it is 6 deficit from 100. So square 6 to get 36, this is the right part of the answer when you want to square 94. Reduce 94 by 6, and you get 88, this is the left part of the answer; so 94^2 = 8836.

So, for 95^2, the deficit is 5, so we have 95-5 giving 90 which is placed to left of 5 square, 9025.

For 91^2, deficit is 9; so 91-9 = 82 is placed to left of 9 square, i.e. 8281.

For 87^2, the deficit is 13, so 13^2 is 169, of which the two digits 69 are written and 1 is carried to left. Meanwhile, the left is 87-13 = 74, add the carried 1, so we get the answer 7569.

Trying 70^2, the deficit is 30, so 30^2 is 900 of which the two zeroes are retained and 9 carried to left. Alnogside, the left is 70-30=40 plus the carried 9, so the answer is 4900.

The Algebra Behind Squaring So Far

2005-10-13T16:01:00.000-07:00

When a number ends in digit 5, e.g. 35,
it is actually 3 x 10 + 5.

So such a number can be of template 10p + 5, where p is any number.
Now squaring this we get 100p^2 + 100p + 25 or
100p(p+1) + 25.

You will appreciate that this is what we have been doing, multiplying p with its one increment; and placing these to left of 25, i.e. effectively multiplying p and p+1 and then multiplying this by 100.

Similarly, a number ending in 25, e.g. 425 is actually 4 x 100 + 25.
So its template is 100q + 25.
Squaring, we get, 10000q^2 + 5000q + 625 or
1000q(10q + 5) + 625.

And again, this is what we have been doing; multiplying q by q placed to left of 5 and then shifting this 3 places, because that is what is meant by writing 625 to the right. See, algebra explains magic so well.

More Squaring

2005-10-12T15:49:00.000-07:00

If you had to square 325, you would have done 32 x 33 and placed this answer to left of 25.

The easier way out here is write 625 as the right part of the answer, and on its left place the product of 35 by 3 (35 is obtained by apending 5 to right of 3). Thus you get 105625.

Sameways, for square of 425, you ndeed 45 x 4 on left of 625, i.e. 180625.

For 525, 55 x5 on left of 625, i.e. 275625

For 725, 75x7 on left of 625, i.e. 525625.

Thus you can square numbers of the style #25.

So 1025 square would be 105x10 to left of 625, i.e. 1050625.

Learn Facts on Squaring

2005-10-11T08:44:00.000-07:00

Let us observe squares of digits ending in 5:

5^2 = 25
15^2 = 225
25^2 = 625
35^2 = 1225
....

What do you note; we always have 25 at the end. For the remaining part to be prefixed to 25, multiply the digit to left of 5 by its one increment.

e.g.. in 35, digit to left of 5 is 3 and its one increment is 3+1 = 4. So 3 x 4 = 12 is prefixing 25.
Easy!

To find square of 65, do one increment on 6, i.e. 7 x 6 = 42; so the answer is 4225.

And in 85, you have 8 x 9 = 72 prefixing 25, i.e. 7225.

In same breath, the square of 115 would be 11 x 12 = 132 prefixing 25, i.e. 13225.

Cyclical Behavior in Powers Last Digit:03

2005-10-10T08:19:00.000-07:00

yesterday you were given these probs:

1) 2^3411
2) 7^201
3) 6^1799

Taking the 1st one. Powers of 2 have last digits 2, 4, 8, 6; repeated in set of 4.
So remainder of 3411 by 4 is 3.
So look at 2^3, which gives 8 is the last digit. This is the answer, i.e.
2 multiplied 3411 times would be a large number like ..................................................8

Second problem has last digits of 7 as 7, 9, 3, 1; again repeated in sets of 4.
So remainder of 201 by 4 is 1, which gives last digit as same in case of 7^1, namely 7.
Thus we can say 7^201 = ...................................................7

Third one is easy, all powers of 6 have last digit 6 only.

It would be useful to keep this table in mind for the last digits of powers:

2, 4, 8, 6.
3, 9, 7, 1.
4, 6.... odd powers 4 and even powers 6.
5...all powers 5.
6...all powers 6.
7, 9, 3, 1.
8, 4, 2, 6.
9, 1... odd powers 9 and even powers 1.
0... all powers 0.

Cyclical Behavior in Powers Last Digit:02

2005-10-09T08:06:00.000-07:00

Continuing from yesterday...

Look, if I ask you to tell me what is the correct response to the question:

2379 x 3456 =
A) 8221822 B) 8221823 C) 8221824 D) 8221825 E) 8221826

you would immediately say the correct answer is Choice C because the last digit of 2379, i.e. 9 and the last digit of 3456, i.e. 6 when multiplied give 54 with 4 as the last digit.

So with same logic, we needn't find fifth power of 3 to find its last digit;
since 3^4 is 81, with last digit 1, the next power has to be given by 1 x 3 = 3.

Same way, 3^6 would be having a last digit obtained by 3 x 3 = 9. This is correct because we know otherwise too that 3^ 6 is 729 with 9 as the last digit.

So this explains the periodic repetition.

Thus to find last digit of 3^2088; simply divide 2086 by 4.
What is the remainder? 2.
So take 3^2 and last digit is 9.
This is the answer.

To repeat, 3^195 has a last digit got like this:
Divide 195 by 4, the remainder is 3. So same last digit as in 3^3, i.e. 7.

Another example, 3^5000.
5000 by 4 gets you a remainder of 0 which is same as 4. So observe the last digit of 3^4, which is 1.

Can you guess the last digit of these:
1) 2^3411
2) 7^201
3) 6^1799

Cyclical Behavior in Powers Last Digit

2005-10-08T08:01:00.000-07:00

Watch these powers of 3:

3^1 = 3
3^2 = 9
3^3 = 27
3^4 = 81
3^5 = 243
3^6 = 729
3^7 = 2187
3^8 = 6561
3^9 = 19683
....
and so on.

Focus on the last digit also called the unit's digit.
You find they are respectively: 3, 9, 7, 1, 3, 9, 7, 1, 3, ...;
so effectively a repetition of the 3-9-7-1 quartet.

Can you think why?
Can you find what would be the last digit of 3^2088 ?

we discuss tomorrow.

Making a Guessing Game:03

2005-10-07T08:34:00.000-07:00

carrying the same thrilling experimentation with numbers:

Let us study a three digit number, say 284.
Now this is actually 100 x 2 + 10 x 8 + 4.

(so mentally you can think it is 110p + 10q 6 r, if the digits are p, q, r from left to right)

On reversing, this would become 482, i.e.
4 x 100 + 10 x 8 + 2.

(so our reverse also becomes 100r + 10q + p).

On subtracting this from the original, we get 284 - 482 = -198
or the difference (difference is without any sign) being 198.

(in our algebra this becomes (100p + 10q + r) - (100r + 10q + p) = 99(p -r))

So if you ask your volunteer to tell you gap of the corner digits of his number (i.e. 2 and 4 of 284 as 2), you would simply multiply this by 99 to get the final answer.

Try this:
Let him think 159. so his gap of corner digits is 8 ( between 9 and 1). You multiply 99 by 8 to get 792.
This is what he would have got by reversing 159 into 951 and subtracting it from 159!

Making a Guessing Game:02

2005-10-06T08:26:00.000-07:00

Let us carry on our hobby of making games.

Let the volunteer think a two digit number.
(Mentally think these are p and q, with p at the tens place and q at the unit's place.
So the worth of the number would be 10p + q.
e.g. 47 is actually 10 x 4 + 7;
and so is 59 = 10 x 5 + 9.)

Ask for this number to be reversed.
(Mentally you get 10 q + p).

On addition, this becomes 11p + 11q, i.e. 11(p+q). Let this be the final answer you are going to tell the volunteer.

So if you are told the sum of the two digits the volunteer has thought, you will multiply it by 11 to give the final answer.

e.g. If the volunteer had thought 28, he will tell you sum of digits is 10 ( 2 + 8).
and you would simply do 11 times this sum to tell him the final answer is 110.

On his part, he would have reversed 28 to get 82, and added 28 & 82 to get the same 110.

Making a Guessing Game

2005-10-05T07:52:00.000-07:00

Let us try making you a magician of sorts.

So you ask your sister to think of a single digit number, viz. 1, 2, 3, 4, 5, 6, 7, 8, or 9;
without telling you.

You ask her to do anything to it, say double.

Then you ask her to add another single digit to the result.

Then you ask her to multiply it by 5.

You ask her to subtract thrice the digit added in second time from the total.

This you ask her to tell you. And lo, you are able to guess her digit.

The Algebra behind this:

Let the digit be p.
Doubling would have made this 2p.
Adding another digit (say q) would have made it 2p + q.
Now, multiplying with 5 would have yielded 10p + 5q.
Subtraction would have left 10p + q.

So if you hear 47, then p is 4 and q is 7.
Similarly, if you finally hear 59, p was 5 and q was 9.

You are encouraged to make couple of such games yourself, keeping track of what is happening algebraically.

PALINDROMES

2005-10-04T19:40:00.000-07:00

A word which spells the same bothways is termed palindrome, e.g. MADAM, MALAYALAM.

Similarly, a number which reads the same left-to-right and right-to-left is a palindrome.
e.g. 747, 236632, 1234321.

You may enjoy the finding that early powers of 11 show palindromes:

11 x 11 = 121
11 x 11 x 11 = 1331
11 x 11 x 11 x 11 = 14641

Is the next one a palindrome too?

37 x 27 = ?

2005-10-02T12:36:00.000-07:00

If you were asked to perform this sum, 37 x 27 = ?,
you would be zapped.

But look, 37 x 3 = 111;
thus 37 x 12 can be seen as 37 x 3 x 4 = 111 x 4 = 444, a neat answer.

Similarily 37 x 27 = 37 x 3 x 9 = 999.

Ain't this lovely.