Bunuel wrote:
\(7^3 + 21^3 =\)
A. \(7^4*2^2\)
B. \(7^3*2^4\)
C. \(7^3*3^3\)
D. \(7^3*3^4\)
E. \(7^4*3^3\)
Method #1: Let us calculate approximate values:
\(7^3 + 21^3 = 7^2 * 7 + 7^3 * 3^3\)
\(= 49 * 7 + 7^3 * 3^3\)
\(≈ 50 * 7 + 7^3 * 3^3\)
\(= 350 + 350 * 27\)
\(≈ 350 + 350 * 30\)
\(= 350 + 10500 = 10850\)
Note: The actual answer is less than 10850
Working with options:
A. \(7^4*2^2 = 49*49*4 ≈ 50*50*4 = 10000\) -- may be possible
B. \(7^3*2^4 = 49*7*16 ≈ 50*7*16 = 50*16*7 = 800 * 7 = 5600\) - too small
C. \(7^3*3^3 = 49*7*27 ≈ 50*7*30 = 350 * 30 = 10500\) -- may be possible
D. \(7^3*3^4 = 49 * 7 * 81 ≈ 50*7*80 = 28000\) -- too large
E. \(7^4*3^3 = 49*49*27 ≈50*50*30 = 75000\) -- too large
Thus, either A or C is possible.
Now, the given question: \(7^3 + 21^3\) is a sum of 2 odd numbers and hence, the result is even
Verifying options A and C:
A: \(7^4*2^2\) is the product of an odd number and an even number => hence, is even ---
matchesC: \(7^3*3^3\) is the product of 2 odd numbers => hence, is odd --- does not match
Answer AAlternate approach: Work with the units digits [Note: The cycle for 3: 3,9,7,1; for 7: 7,9,3,1; 2: 2,4,8,6]:
Units digit of the question: \(7^3 + 21^3\) = Units digit of \(3 + 1 = 4\)
Working with the options:
A. Units digit of \(7^4*2^2\) = Units digit of \(1 * 4 = 4\) ---
matches B. Units digit of \(7^3*2^4\) = Units digit of \(3 * 6 = 8\) --- does not match
C. Units digit of \(7^3*3^3\) = Units digit of \(3 * 7 = 1\) --- does not match
D. Units digit of \(7^3*3^4\) = Units digit of \(3 * 1 = 3\) --- does not match
E. Units digit of \(7^4*3^3\) = Units digit of \(1 * 7 = 7\) --- does not match
Answer AApproach #3: Those who are good with simplification, the best way of solving is as follows:
\(7^3 + 21^3 = 7^3 + 7^3 * 3^3\)
\(= 7^3 * (1 + 3^3)\)
\(= 7^3 * (1 + 27)\)
\(= 7^3 * 28\)
\(= 7^3 * 7 * 2^2\)
\(= 7^4 * 2^2\)
Answer A _________________
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