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Each of the sets F_0, F_1,
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28 Feb 2025, 00:36
28 Feb 2025, 00:36
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50% (01:31) wrong based on 8 sessions
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Each of the sets $\(\mathrm{F}_0, \mathrm{~F}_1, \mathrm{~F}_2, \mathrm{~F}_3, \ldots \ldots \ldots . . . . . . . . . \mathrm{F}_9\)$ satisfy the condition that each of them contain all the integers that are ending with the digit of their set number. For example $\(\mathrm{F}_5\)$ has integers $\(5,15,25,35\)$, ....... etc. The cubes of all the numbers of set $\(\mathrm{F}_7\)$ are present in which of the following set?
(A) None
(B) $\(\mathrm{F}_2\)$
(C) $\(\mathrm{F}_3\)$
(D) $\(\mathrm{F}_7\)$
(E) $\(\mathrm{F}_9\)$
(A) None
(B) $\(\mathrm{F}_2\)$
(C) $\(\mathrm{F}_3\)$
(D) $\(\mathrm{F}_7\)$
(E) $\(\mathrm{F}_9\)$
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Each of the sets F_0, F_1,
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02 Mar 2025, 13:03
02 Mar 2025, 13:03
1
\(F_7\) contains 7, 17, 27, etc.
The cube of 7 = 7 * 7 * 7 = 343
You could actually stop here, save time, and determine that the set must be \(F_3\), since 343 ends in 3.
No other set aside from \(F_3\) would contain the integer 343 since all of the integers in all of the other sets end with a digit other than 3.
Just to further prove the point though:
The cube of 17 = 17 * 17 * 17 = 4913
The cube of 27 = 27 * 27 * 27 = 19683
Again, the cubes end in the digit 3, so they are contained in \(F_3\)
The cube of 7 = 7 * 7 * 7 = 343
You could actually stop here, save time, and determine that the set must be \(F_3\), since 343 ends in 3.
No other set aside from \(F_3\) would contain the integer 343 since all of the integers in all of the other sets end with a digit other than 3.
Just to further prove the point though:
The cube of 17 = 17 * 17 * 17 = 4913
The cube of 27 = 27 * 27 * 27 = 19683
Again, the cubes end in the digit 3, so they are contained in \(F_3\)
gmatclubot
Each of the sets F_0, F_1, [#permalink]
02 Mar 2025, 13:03
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