# Remainder of large multiples

• June 24th 2005, 12:07 AM
Help-hopeless with maths
Question 1
a/ Find the remainder when 7100 is divided by 25
b/ Find the remainder when 2603 * 3201 is divided by 25
c/ Hence show that 29 * 7100 + 54 * 2603 * 3201 is divisible by 25

Question 2
In a triangle ABC, BAC = 35°. D is a point on AB such that AD=BC and ADC=110°. E is a point on AC such that BE=DC. What is the size of ABE?

:confused: :confused:
• June 24th 2005, 01:57 AM
tbsmith
1a) Since 25 divides evenly into 100, and 100 divides evenly into 7100, there must be a zero remainder

1b) 2603*3201 = (2600+3)*(3200+1) = (2600*3200 + 2600 + 3*3200 + 3) = 3 (mod 25)

1c) My Maple says:

29 * 7100 + 54 * 2603 * 3201 mod 25 = 12

So you've typed it in wrong.

In 2) I think ABE=0, because you need B=D and E=A to satisfy the conditions of the problem.
• June 24th 2005, 03:42 AM
ticbol
1.)
Without going into details, how can
29*7100 +54*2603*3201
be divisible by 25 when the units digit of the sum is 2?

29*71200 = _ _ ..._ _ _0
plus
54*2603*3201 = _ _ ..._ _ _2
-------------------------
Total = _ _ ..._ _ _2 <----ends in 2.

A number can possibly be divisible by 25 if this number ends in 0 or 5.
(That does not mean, though, that the 0- or 5-ending number is always be divisible by 25.)

=======================
2.)
Draw the figure on paper.

angle DAC is the same as angle BAC. .....[= 35deg ----given.]
angle ADC = 110 deg .....[given]
So, angle DCA = 180 -35 -110 = 35 deg also
That means angle DAC = angle DCA

-------
In triangle DCB:

So, AD = DC = BC
Hence, triangle DCB is isosceles too.

angle BDC = 180 -110 = 70 deg
So, angle DBC = 70 deg too.
Then, angle DCB = 180 -70 -70 = 40 deg

-----------
In triangle EBC:

AD = DC = BC .....[found, or arrived at]
BE = DC .....[given]
So, BE = BC
Hence, triangle EBC is isosceles too.

angle DCB = 40 deg ....[found]
angle DCA = 35 deg ....[found]
So, angle ECB = 35+40 = 75 deg
Hence, angle CEB = 75 deg too.
Then, angle EBC = 180 -75 -75 = 30 deg

---------
In triangle ABC:

angle DBC = 70 deg ....[found]
angle EBC = 30 deg ....[found]
Therefore, angle ABE = 70 -30 = 40 deg .....answer.
• June 24th 2005, 02:04 PM
Sorry typed wrong
a/ Find the remainder when 7100 is divided by 25
b/ Find the remainder when 2603 * 3201 is divided by 25
c/ Hence show that 29 * 7100 + 54 * 2603 * 3201 is divisible by 25

:eek: Sorry its not that it is this, for some reason I can't do power of numbers on the computer:

Question 1
a/ Find the remainder when 7 to the power of 100 is divided by 25
b/ Find the remainder when 2 to the power of 603* 3 to the power of 201 is divided by 25
c/ Hence show that 29 * (7 to the power of 100) + 54 * (2 to the power of 603) * (3 to the power of 201) is divisible by 25
• June 24th 2005, 02:30 PM
Math Help
Power notation
To show powers we use ^ (shift+6) like 2^100

or learn to use latex click on the following equation to see what I mean:

$2^{100}$
• June 25th 2005, 03:41 PM
ticbol
The Question 1
Zeez, I like this.
If I only knew before that your intention for Question 1 is like that, then I should not have touched your posting because the ways to solve your Question 1 were beyond me. I mean I didn't know how to solve Question 1.

That is why it is important, for me, that you separate questions, or you post questions separately, if the two or more questions are difficult to solve each. This way I can answer the questions that are within my reach, and I will just leave those that are beyond me. Otherwise, I will not answer your combined posting because I cannot solve one or two of the questions in it.

I did not study Modular Math. It was not taught to us in school back then. It was not included in the Math taught us in school back then.
So I read about Modular Math in the Internet a while back, and I think I understand it enough to answer your Question 1 here.

I may not be that good yet. This is my very first Modular Math answer, ever.
I will improve, in time, if I can practice more.

------------
>>>a/ Find the remainder when 7 to the power of 100 is divided by 25.

We use Modular Math, or modular arithmetic, or "clock arithmetic", or congruence arithmetic, or whatever it is called.
Since we are dividing by 25, then we use "mod 25" or with respect to modulus 25.
Our set of numbers here are from 0 to 24 ----these are the numbers inside the modulus 25.

The remainder when 7^100 is divided by 25 can be written as (7^100) mod 25.
That is equivalent to (7 mod 25)^100.
Then, 7 divided by 25. Cannot be. 7 is smaller than 25. We need a number from 7^100 that is bigger than 25. We choose 7^2 = 49.

Then, 7^100 = 7^(2*50) = (7^2)^50 = (49)^50 ...[49/25 = 1 rem 24]
= (24)^50
We are not done yet,
= (24)^(2*25) = (24^2)^25 = (576)^25 mod 25 ....[576/25 = 23 rem 1]
= (1)^25

That means,
7^100 mod 25 = 1, or = 1 (mod 25).
or, 7^100 divided 25 = (some integer multiple of 25) + remainder 1.

-------------------
>>>b/ Find the remainder when 2 to the power of 603* 3 to the power of 201 is divided by 25.

Still in mod 25.

[2^(603) * 3^(201)] = [2^(3*201) * 3^(201)] = [(2^3)^(201) * 3^(201)] = [8^(201) * 3^(201)] = [(8*3)^(201)] = 24^201
= 24^(1 +200) = 24[24^200]
= 24[(24^2)^100] = 24[(576)^100] = 24[1^100] = 24(1) = 24 ----the remainder, answer

Meaning,
[2^(603) * 3^(201)] = 24 (mod 25).

------------------
>>>c/ Hence show that 29 * (7 to the power of 100) + 54 * (2 to the power of 603) * (3 to the power of 201) is divisible by 25

In mod 25,
29[7^100] +54[2^603 * 3^201] = 0?

---if divisible by 25, then there is no remainder. Or, remainder = 0.

29 and 54 are outside of the numbers that are inside modulus 25, so we need to put 29 and 54 into mod 25.

29[7^100] +54[2^603 * 3^201]
Substitutions, see parts a and b above,
= 29[1] +54[24]
= [4][1] +[4][24] ----here, 29 = 4, and 54 = 4 also ---(in mod 25).

From here we can go many ways.

One way is
= 4(1) +4(-1) -----24 is -1 in mod 25.
= 4 -4
= 0
Therefore, shown.

Another way,
= 4(1) +4(24)
= 4 +96 .....[96/25 = 3 +rem 21] ....***(see note below)
= 4 +21
= 25
= 0
Therefore, shown.

***
Note, 4+96 = 100, which is divisible by 25. Or, 100 = 0 (mod 25).

================
I am sure the are other ways to solve this question in modular math. Mine above could be crude, but it gets the answer.
• June 27th 2005, 12:17 AM