Question:
A positive whole number \( M \) less than 100 is represented in base 2 notation, base 3 notation, and base 5 notation. It is found that in all three cases the last digit is 1, while in exactly two out of the three cases the leading digit is 1. Then \( M \) equals
A positive whole number \( M \) less than 100 is represented in base 2 notation, base 3 notation, and base 5 notation. It is found that in all three cases the last digit is 1, while in exactly two out of the three cases the leading digit is 1. Then \( M \) equals
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A last digit of 1 in bases 2, 3 and 5 means that \( M - 1 \) is divisible by all of 2, 3 and 5. List the few possible values of \( M \), then write each in the three bases.
Updated On: Nov 7, 2026
- 31
- 63
- 75
- 91
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Question:
We need a whole number \(M < 100\) whose last digit is \(1\) in base 2, base 3 and base 5, and whose leading (first) digit is \(1\) in exactly two of these three bases.
Step 2: Use the last digit condition:
The last digit of \(M\) in base \(k\) is the remainder when \(M\) is divided by \(k\). So the remainder is \(1\) for \(k = 2, 3, 5\). This means \(M - 1\) is divisible by \(2\), \(3\) and \(5\), so by their lcm \(30\).
\[ M = 30t + 1 \Rightarrow M \in \{1, 31, 61, 91\} \quad (M < 100) \]
The value \(M = 1\) is written as "1" in every base, so all three leading digits are \(1\), not exactly two. We drop it. The candidates are \(31\), \(61\), \(91\).
Step 3: Convert the candidates:
\(31\): base 2 is \(11111\) (leading 1); base 3 is \(1011\) because \(27 + 3 + 1 = 31\) (leading 1); base 5 is \(111\) because \(25 + 5 + 1 = 31\) (leading 1). All three leading digits are \(1\), which is too many.
\(61\): base 2 is \(111101\) (leading 1); base 3 is \(2021\) because \(54 + 6 + 1 = 61\) (leading 2); base 5 is \(221\) because \(50 + 10 + 1 = 61\) (leading 2). Only one leading digit is \(1\), which is too few.
\(91\): base 2 is \(1011011\) because \(64 + 16 + 8 + 2 + 1 = 91\) (leading 1); base 3 is \(10101\) because \(81 + 9 + 1 = 91\) (leading 1); base 5 is \(331\) because \(75 + 15 + 1 = 91\) (leading 3). Exactly two leading digits are \(1\). All three last digits are \(1\) as well.
Step 4: Check the options:
Option 1 (\(31\)) gives three leading ones, so it fails. Option 2 (\(63\)) gives last digit \(0\) in base 3 (\(63 = 2100_3\)), so it fails. Option 3 (\(75\)) ends in \(0\) in base 3 and base 5 (\(75 = 2210_3 = 300_5\)), so it fails.
Option 4 (\(91\)) satisfies every condition.
Final Answer:
The number is \(M = 91\), which is option 4. \[ \boxed{91 \text{ (Option 4)}} \]
We need a whole number \(M < 100\) whose last digit is \(1\) in base 2, base 3 and base 5, and whose leading (first) digit is \(1\) in exactly two of these three bases.
Step 2: Use the last digit condition:
The last digit of \(M\) in base \(k\) is the remainder when \(M\) is divided by \(k\). So the remainder is \(1\) for \(k = 2, 3, 5\). This means \(M - 1\) is divisible by \(2\), \(3\) and \(5\), so by their lcm \(30\).
\[ M = 30t + 1 \Rightarrow M \in \{1, 31, 61, 91\} \quad (M < 100) \]
The value \(M = 1\) is written as "1" in every base, so all three leading digits are \(1\), not exactly two. We drop it. The candidates are \(31\), \(61\), \(91\).
Step 3: Convert the candidates:
\(31\): base 2 is \(11111\) (leading 1); base 3 is \(1011\) because \(27 + 3 + 1 = 31\) (leading 1); base 5 is \(111\) because \(25 + 5 + 1 = 31\) (leading 1). All three leading digits are \(1\), which is too many.
\(61\): base 2 is \(111101\) (leading 1); base 3 is \(2021\) because \(54 + 6 + 1 = 61\) (leading 2); base 5 is \(221\) because \(50 + 10 + 1 = 61\) (leading 2). Only one leading digit is \(1\), which is too few.
\(91\): base 2 is \(1011011\) because \(64 + 16 + 8 + 2 + 1 = 91\) (leading 1); base 3 is \(10101\) because \(81 + 9 + 1 = 91\) (leading 1); base 5 is \(331\) because \(75 + 15 + 1 = 91\) (leading 3). Exactly two leading digits are \(1\). All three last digits are \(1\) as well.
Step 4: Check the options:
Option 1 (\(31\)) gives three leading ones, so it fails. Option 2 (\(63\)) gives last digit \(0\) in base 3 (\(63 = 2100_3\)), so it fails. Option 3 (\(75\)) ends in \(0\) in base 3 and base 5 (\(75 = 2210_3 = 300_5\)), so it fails.
Option 4 (\(91\)) satisfies every condition.
Final Answer:
The number is \(M = 91\), which is option 4. \[ \boxed{91 \text{ (Option 4)}} \]
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