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16 Sep 2014, 01:30
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C
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Difficulty:
75%
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Question Stats:
50% (01:48) correct
50%
(02:15)
wrong
based on 219
sessions
History
Date
Time
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If the least common multiple of a positive integer \(x\), \(4^3\) and \(6^5\) is \(6^6\), then how many values can \(x\) take?
A. \(1\)
B. \(6\)
C. \(7\)
D. \(30\)
E. \(36\)
A. \(1\)
B. \(6\)
C. \(7\)
D. \(30\)
E. \(36\)
16 Sep 2014, 01:30
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Official Solution:
If the least common multiple of a positive integer \(x\), \(4^3\) and \(6^5\) is \(6^6\), then how many values can \(x\) take?
A. \(1\)
B. \(6\)
C. \(7\)
D. \(30\)
E. \(36\)
We are given that \(6^6=2^{6}*3^{6}\) is the least common multiple (LCM) of three numbers:
\(x\);
\(4^3=2^6\);
\(6^5 = 2^{5}*3^5\);
First notice that \(x\) cannot include any prime factors other than 2 and/or 3 as the LCM is exclusively composed of these primes.
Next, since the exponent of 3 in the LCM is greater than the exponent of 3 in the other two numbers, \(x\) must have \(3^6\) as a factor. Otherwise, the factor \(3^{6}\) wouldn't appear in the LCM.
Furthermore, \(x\) can include the prime number 2 raised to any power from 0 to 6, inclusive (the LCM limits the power of 2 in \(x\) to 6, so it cannot be any higher).
Therefore, \(x\) can take on a total of 7 distinct values, namely:
\(3^6\);
\(2*3^6\);
\(2^2*3^6\);
\(2^3*3^6\);
\(2^4*3^6\);
\(2^5*3^6\);
\(2^6*3^6\).
Answer: C
If the least common multiple of a positive integer \(x\), \(4^3\) and \(6^5\) is \(6^6\), then how many values can \(x\) take?
A. \(1\)
B. \(6\)
C. \(7\)
D. \(30\)
E. \(36\)
We are given that \(6^6=2^{6}*3^{6}\) is the least common multiple (LCM) of three numbers:
\(x\);
\(4^3=2^6\);
\(6^5 = 2^{5}*3^5\);
First notice that \(x\) cannot include any prime factors other than 2 and/or 3 as the LCM is exclusively composed of these primes.
Next, since the exponent of 3 in the LCM is greater than the exponent of 3 in the other two numbers, \(x\) must have \(3^6\) as a factor. Otherwise, the factor \(3^{6}\) wouldn't appear in the LCM.
Furthermore, \(x\) can include the prime number 2 raised to any power from 0 to 6, inclusive (the LCM limits the power of 2 in \(x\) to 6, so it cannot be any higher).
Therefore, \(x\) can take on a total of 7 distinct values, namely:
\(3^6\);
\(2*3^6\);
\(2^2*3^6\);
\(2^3*3^6\);
\(2^4*3^6\);
\(2^5*3^6\);
\(2^6*3^6\).
Answer: C
dcummins
Joined: 14 Feb 2017
Last visit: 08 May 2025
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Given Kudos: 368
Location: Australia
Concentration: Technology, Strategy
Schools: INSEAD Sept'21 Intake (WA$) Melbourne (A$) AGSM (A$) Kellogg '23 (WL) Said '22 (A$) CBS '23 (D) Booth '23 (D) Anderson '23 (WL) Cambridge '22 (WL) LBS '23 (D) Stern '24 (WA)
GMAT 1: 560 Q41 V26
GMAT 2: 550 Q43 V23
GMAT 3: 650 Q47 V33
GMAT 4: 650 Q44 V36
GMAT 5: 600 Q38 V35
GMAT 6: 710 Q47 V41
WE:Management Consulting (Consulting)
Products:
Schools: INSEAD Sept'21 Intake (WA$) Melbourne (A$) AGSM (A$) Kellogg '23 (WL) Said '22 (A$) CBS '23 (D) Booth '23 (D) Anderson '23 (WL) Cambridge '22 (WL) LBS '23 (D) Stern '24 (WA)
GMAT 6: 710 Q47 V41
Posts: 1,070
28 Apr 2019, 20:40
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If the least common multiple of a positive integer \(x\), \(4^3\) and \(6^5\) is \(6^6\). Then \(x\) can take how many values?
We determine the LCM by counting the highest number of powers of a given prime that is repeated across the set * multiplied by non-repeated powers
If the LCM is \(6^6\) then only the primes 2 and 3 are present and the highest power of 3 must be contained within \(x\) itself since:
\(4^3 = 2^6\) does not contain \(3^6\)
and
\(6^5 = (3^5 * 2^5)\) does not contain \(3^6\)
Thus x must contain 3^6
1 solution is \(3^6\) (as determined already)..but what else can x contain?
X can contain contain primes found within the LCM, so it can contain multiples of 2 up to the upper limit, which is given by the LCM of \(6^6 (2^6)\)
Solution 2 = \(3^6 * 2^1\)
Solution 3 = \(3^6 * 2^2\)
Solution 4 = \(3^6 * 2^3\)
Solution 5 = \(3^6 * 2^4\)
Solution 6 = \(3^6 * 2^5\)
Solution 7 = \(3^6 * 2^6\)
We determine the LCM by counting the highest number of powers of a given prime that is repeated across the set * multiplied by non-repeated powers
If the LCM is \(6^6\) then only the primes 2 and 3 are present and the highest power of 3 must be contained within \(x\) itself since:
\(4^3 = 2^6\) does not contain \(3^6\)
and
\(6^5 = (3^5 * 2^5)\) does not contain \(3^6\)
Thus x must contain 3^6
1 solution is \(3^6\) (as determined already)..but what else can x contain?
X can contain contain primes found within the LCM, so it can contain multiples of 2 up to the upper limit, which is given by the LCM of \(6^6 (2^6)\)
Solution 2 = \(3^6 * 2^1\)
Solution 3 = \(3^6 * 2^2\)
Solution 4 = \(3^6 * 2^3\)
Solution 5 = \(3^6 * 2^4\)
Solution 6 = \(3^6 * 2^5\)
Solution 7 = \(3^6 * 2^6\)
