CAT 2025 Quant was dominated by Algebra followed by Arithmetic. In Arithmetic, the questions were dominated by topics like Speed-time-distance, Mixture and Alligations. This year, there was a surprise. The questions from Geometry were relatively on the lower side as compared to the previous years. There were 8 TITA Qs this year. Overall this section was at a medium level of difficulty.
Question 11 : The number of divisors of \( \left( 2 ^ { 6 } \times 3 ^ { 5 } \times 5 ^ { 3 } \times 7 ^ { 2 } \right) \), which are of the form \( ( 3 r + 1 ) \), where \( r \) is a non-negative integer, is
Exclude factor 3
No multiple of 3 can be of the form \(3r+1\). Thus, power of 3 must be \(3^0\).
(1 possibility.)
Analyze powers of 7
\(7\equiv1\pmod3\), so any power of 7 leaves remainder 1.
\(c\in\{0,1,2\}\Rightarrow3\) choices for powers of 7.
Determine valid combinations of powers of 2 and 5
\(2\equiv2\pmod3\) and \(5\equiv2\pmod3\).
\(2^a\pmod3=1\) when \(a\) is even:
\(a\in\{0,2,4,6\}\) (4 choices).
\(2^a\pmod3=2\) when \(a\) is odd:
\(a\in\{1,3,5\}\) (3 choices).
\(5^b\pmod3=1\) when \(b\) is even:
\(b\in\{0,2\}\) (2 choices).
\(5^b\pmod3=2\) when \(b\) is odd:
\(b\in\{1,3\}\) (2 choices).
To get product \(\equiv1\pmod3\):
Case A (Both Remainder 1): even \(a\)\(\times\)even \(b=4\times2=8\) pairs
Case B (Both Remainder 2): odd \(a\)\(\times\)odd \(b=3\times2=6\) pairs
Total valid \(a,b\) pairs \(=8+6=14\).
Compute total divisors
Total Divisors \(=14\times3=42\)
The question is " The number of divisors of \( \left( 2 ^ { 6 } \times 3 ^ { 5 } \times 5 ^ { 3 } \times 7 ^ { 2 } \right) \), which are of the form \( ( 3 r + 1 ) \), where \( r \) is a non-negative integer, is "
Choice 1 is the correct answer.
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