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Find the greatest integer:
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Updated on: 29 Oct 2017, 09:26
Updated on: 29 Oct 2017, 09:26
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60% (01:03) correct
39% (01:55) wrong based on 28 sessions
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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?
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?
Re: Find the greatest integer:
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01 Nov 2017, 19:12
01 Nov 2017, 19:12
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Expert Reply
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?
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
Re: Find the greatest integer:
[#permalink]
14 Jun 2019, 06:57
14 Jun 2019, 06:57
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?
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 ?
