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TAGS: Intern Joined: 01 Oct 2017
Posts: 3
Followers: 0

Kudos [?]: 2  , given: 1

1
KUDOS 00:00

Question Stats: 75% (01:07) correct 25% (00:24) wrong based on 8 sessions
Find the greatest integer

a) $$10^{10} + 2^{100}$$

b) $$100^{10} + 2^{10}$$

c) $$(100 + 2 ) ^{10}$$

Can anyone suggest me the efficient way to solve this question?
[Reveal] Spoiler: OA

Last edited by Carcass on 29 Oct 2017, 09:26, edited 1 time in total.
Edited by Carcass Supreme Moderator
Joined: 01 Nov 2017
Posts: 370
Followers: 8

Kudos [?]: 138  , given: 4

Re: Find the greatest integer: [#permalink]
2
KUDOS
Expert's post
sandesh10 wrote:
Find the greatest integer

a) $$10^{10} + 2^{100}$$

b) $$100^{10} + 2^{10}$$

c) $$(100 + 2 ) ^{10}$$

Can anyone suggest me the efficient way to solve this question?

Hi...

The easiest way would be..
Compare a and B..

a) $$10^{10} + 2^{100}$$

b) $$100^{10} + 2^{10}= (10^{10})^{10}+2^{10}=10^{100}+2^{10}$$

Now when you compare two 10^{100} is way GREATER than 2^{100} as compared to 10^{10} and 2^{10} ..

Otherwise a exact way would be...

a) $$10^{10} + 2^{100}=10^{10}+(2^{10})^10=10^{10}+1024^{10}=10^{10}+10^{30}~10^30$$

b) $$100^{10} + 2^{10}=10^{100}+10^3$$

SO B is clearly the bigger one

Now let's compare B and C

c) $$(100 + 2 ) ^{10}$$
When you expand it 100^{10}+100^9*2^1+.....+2^10

So here C is bigger..

Also

b) $$100^{10} + 2^{10}~100^{10}$$

c) $$(100 + 2 ) ^{10}=102^{10}$$

C wins
_________________

Some useful Theory.
1. Arithmetic and Geometric progressions : https://greprepclub.com/forum/progressions-arithmetic-geometric-and-harmonic-11574.html#p27048
2. Effect of Arithmetic Operations on fraction : https://greprepclub.com/forum/effects-of-arithmetic-operations-on-fractions-11573.html?sid=d570445335a783891cd4d48a17db9825
3. Remainders : https://greprepclub.com/forum/remainders-what-you-should-know-11524.html
4. Number properties : https://greprepclub.com/forum/number-property-all-you-require-11518.html
5. Absolute Modulus and Inequalities : https://greprepclub.com/forum/absolute-modulus-a-better-understanding-11281.html

Manager Joined: 08 Dec 2018
Posts: 94
Followers: 0

Kudos [?]: 37 , given: 30

Re: Find the greatest integer: [#permalink]
chetan2u wrote:
sandesh10 wrote:
Find the greatest integer

a) $$10^{10} + 2^{100}$$

b) $$100^{10} + 2^{10}$$

c) $$(100 + 2 ) ^{10}$$

Can anyone suggest me the efficient way to solve this question?

Hi...

The easiest way would be..
Compare a and B..

a) $$10^{10} + 2^{100}$$

b) $$100^{10} + 2^{10}= (10^{10})^{10}+2^{10}=10^{100}+2^{10}$$

Now when you compare two 10^{100} is way GREATER than 2^{100} as compared to 10^{10} and 2^{10} ..

Otherwise a exact way would be...

a) $$10^{10} + 2^{100}=10^{10}+(2^{10})^10=10^{10}+1024^{10}=10^{10}+10^{30}~10^30$$

b) $$100^{10} + 2^{10}=10^{100}+10^3$$

SO B is clearly the bigger one

Now let's compare B and C

c) $$(100 + 2 ) ^{10}$$
When you expand it 100^{10}+100^9*2^1+.....+2^10

So here C is bigger..

Also

b) $$100^{10} + 2^{10}~100^{10}$$

c) $$(100 + 2 ) ^{10}=102^{10}$$

C wins

for option b,

100^10+2^10=> (10^2)^10+2^10=>10^20+2^10.
Isn't it ? Re: Find the greatest integer:   [#permalink] 14 Jun 2019, 06:57
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