What are the first six digits of 1/9+1/99+1/999, if expressed as a decimal?
a .122123
b .122213
c .123123
d .123212
e .123321
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ganeshrkamath
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a/9 = 0.aaaaa...kop wrote:What are the first six digits of 1/9+1/99+1/999, if expressed as a decimal?
a .122123
b .122213
c .123123
d .123212
e .123321
ab/99 = 0.abababab...
abc/999 = 0.abcabcabc...
(a,b,c are digits)
1/9 = 0.111111...
01/99 = 0.010101...
001/999 = 0.001001...
Sum = 0.122213...
Choose B
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Mathsbuddy
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In the answer list the 1st 2 digits are all the same, so we only need to consider the 3rd and 4th digits to identify which is correct.
Knowing:
1/9 -> decimal positions of 1 are multiples of 1
1/99 -> decimal positions of 1 are multiples of 2
1/999 -> decimal positions of 1 are multiples of 3
gives us (3rd, 4th) digits of (1+1, 1+1) = (2,2)
Therefore answer B.
Knowing:
1/9 -> decimal positions of 1 are multiples of 1
1/99 -> decimal positions of 1 are multiples of 2
1/999 -> decimal positions of 1 are multiples of 3
gives us (3rd, 4th) digits of (1+1, 1+1) = (2,2)
Therefore answer B.
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sanju09
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1/9 =0 .111111kop wrote:What are the first six digits of 1/9+1/99+1/999, if expressed as a decimal?
a .122123
b .122213
c .123123
d .123212
e .123321
1/99 = 0.010101
1/999 = 0.001001
ADD
3 in the last place, eliminate D and E, 1 in fifth place, eliminate A and C. We can stop and take the leftover.
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Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
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Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
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sanju09
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There must be something wrong with this website, a little edit work in post creates two posts when submitted.
The mind is everything. What you think you become. -Lord Buddha
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
Hi,Mathsbuddy wrote:In the answer list the 1st 2 digits are all the same, so we only need to consider the 3rd and 4th digits to identify which is correct.
Knowing:
1/9 -> decimal positions of 1 are multiples of 1
1/99 -> decimal positions of 1 are multiples of 2
1/999 -> decimal positions of 1 are multiples of 3
gives us (3rd, 4th) digits of (1+1, 1+1) = (2,2)
Therefore answer B.
Can you elaborate more on your steps please ?
More on "decimal positions of 1 are multiples of 1"etc.
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Hi buoyant,
Mathsbuddy is referring to the patterns that exist in those 3 fractions:
1/9 = .111111
2/9 = .222222
3/9 = .333333
etc.
1/99 = .010101
2/99 = .020202
3/99 = .030303
11/99 = .111111
etc.
1/999 = .001001
2/999 = .002002
45/999 = .045045
217/999 = .217217
etc.
By adding up 1/9 + 1/99 + 1/999 =
.111111 + .010101 + .001001
GMAT assassins aren't born, they're made,
Rich
Mathsbuddy is referring to the patterns that exist in those 3 fractions:
1/9 = .111111
2/9 = .222222
3/9 = .333333
etc.
1/99 = .010101
2/99 = .020202
3/99 = .030303
11/99 = .111111
etc.
1/999 = .001001
2/999 = .002002
45/999 = .045045
217/999 = .217217
etc.
By adding up 1/9 + 1/99 + 1/999 =
.111111 + .010101 + .001001
GMAT assassins aren't born, they're made,
Rich
