CMAT 2026: Number System
Number System Quick Review
Cyclicity of 4:
\(4^1 = 4\) (Odd power ends in 4)
\(4^2 = 6\) (Even power ends in 6)
\(4^1 = 4\) (Odd power ends in 4)
\(4^2 = 6\) (Even power ends in 6)
LCM Application:
To find when multiple events coincide, find the LCM of their time intervals.
To find when multiple events coincide, find the LCM of their time intervals.
Telescoping Sums:
Look for terms like \(\frac{1}{n^2} – \frac{1}{(n+1)^2}\) where middle terms cancel out.
Look for terms like \(\frac{1}{n^2} – \frac{1}{(n+1)^2}\) where middle terms cancel out.
The digit in the unit’s place of the resulting number of the expression \((234)^{100} + (234)^{101}\) is:
Correct Option: 4 (Value: 0)
Rationale: We look at the cyclicity of the base digit (4).
– For \((234)^{100}\): Power 100 is even, so the unit digit is 6.
– For \((234)^{101}\): Power 101 is odd, so the unit digit is 4.
Sum = \(6 + 4 = 10\). The resulting unit digit is 0.
Why other options wrong: Ignoring cyclicity rules or miscalculating the sum of digits.
Difficulty: Easy
Rationale: We look at the cyclicity of the base digit (4).
– For \((234)^{100}\): Power 100 is even, so the unit digit is 6.
– For \((234)^{101}\): Power 101 is odd, so the unit digit is 4.
Sum = \(6 + 4 = 10\). The resulting unit digit is 0.
Why other options wrong: Ignoring cyclicity rules or miscalculating the sum of digits.
Difficulty: Easy
Four different electronic devices make a beep after every 30 minutes, 1 hour, 1 1/2 hour and 1 hour 45 minutes, respectively. All these devices beeped together at 12 noon. They will again beep together at:
Correct Option: 4 (9 A.M.)
Rationale: Convert intervals to minutes: 30, 60, 90, and 105.
LCM(30, 60, 90, 105) = \(2^2 \times 3^2 \times 5 \times 7 = 1260\) minutes.
\(1260 \text{ minutes} = 21 \text{ hours}\).
12 Noon + 21 hours = 9 A.M. (next day).
Why other options wrong: Miscalculating the LCM or adding the hours incorrectly to 12 noon.
Difficulty: Moderate
Rationale: Convert intervals to minutes: 30, 60, 90, and 105.
LCM(30, 60, 90, 105) = \(2^2 \times 3^2 \times 5 \times 7 = 1260\) minutes.
\(1260 \text{ minutes} = 21 \text{ hours}\).
12 Noon + 21 hours = 9 A.M. (next day).
Why other options wrong: Miscalculating the LCM or adding the hours incorrectly to 12 noon.
Difficulty: Moderate
The value of the following sum of terms:
\(\frac{5}{2^2 \cdot 3^2} + \frac{7}{3^2 \cdot 4^2} + \frac{9}{4^2 \cdot 5^2} + \frac{11}{5^2 \cdot 6^2} + \frac{13}{6^2 \cdot 7^2} + \frac{15}{7^2 \cdot 8^2} =\)
\(\frac{5}{2^2 \cdot 3^2} + \frac{7}{3^2 \cdot 4^2} + \frac{9}{4^2 \cdot 5^2} + \frac{11}{5^2 \cdot 6^2} + \frac{13}{6^2 \cdot 7^2} + \frac{15}{7^2 \cdot 8^2} =\)
Correct Option: 2 (15/64)
Rationale: This is a telescoping series. General term: \(\frac{1}{n^2} – \frac{1}{(n+1)^2}\).
Series: \((\frac{1}{4} – \frac{1}{9}) + (\frac{1}{9} – \frac{1}{16}) + \dots + (\frac{1}{49} – \frac{1}{64})\).
Result: \(\frac{1}{4} – \frac{1}{64} = \frac{16-1}{64} = \frac{15}{64}\).
Why other options wrong: Calculation error using wrong bases or incorrect subtraction.
Difficulty: Moderate
Rationale: This is a telescoping series. General term: \(\frac{1}{n^2} – \frac{1}{(n+1)^2}\).
Series: \((\frac{1}{4} – \frac{1}{9}) + (\frac{1}{9} – \frac{1}{16}) + \dots + (\frac{1}{49} – \frac{1}{64})\).
Result: \(\frac{1}{4} – \frac{1}{64} = \frac{16-1}{64} = \frac{15}{64}\).
Why other options wrong: Calculation error using wrong bases or incorrect subtraction.
Difficulty: Moderate
The value of expression \(\frac{\sqrt{\sqrt{5}+2} + \sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}\) is:
Correct Option: 2 (\(\sqrt{2}\))
Rationale: Let \(X = \frac{N}{D}\). Square the numerator \(N\):
\(N^2 = (\sqrt{5}+2) + (\sqrt{5}-2) + 2\sqrt{(\sqrt{5}+2)(\sqrt{5}-2)}\)
\(N^2 = 2\sqrt{5} + 2\sqrt{5-4} = 2\sqrt{5} + 2 = 2(\sqrt{5}+1)\).
Since \(D^2 = \sqrt{5}+1\), then \(X^2 = \frac{N^2}{D^2} = 2\).
Thus, \(X = \sqrt{2}\).
Why other options wrong: Ignoring interaction terms or miscalculating the difference of squares.
Difficulty: Hard
Rationale: Let \(X = \frac{N}{D}\). Square the numerator \(N\):
\(N^2 = (\sqrt{5}+2) + (\sqrt{5}-2) + 2\sqrt{(\sqrt{5}+2)(\sqrt{5}-2)}\)
\(N^2 = 2\sqrt{5} + 2\sqrt{5-4} = 2\sqrt{5} + 2 = 2(\sqrt{5}+1)\).
Since \(D^2 = \sqrt{5}+1\), then \(X^2 = \frac{N^2}{D^2} = 2\).
Thus, \(X = \sqrt{2}\).
Why other options wrong: Ignoring interaction terms or miscalculating the difference of squares.
Difficulty: Hard
