Question:
The product of the digits of a three-digit number is 70. The sum of the digits of this three-digit number is _____
The product of the digits of a three-digit number is 70. The sum of the digits of this three-digit number is _____
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Break 70 into three factors that are each a single digit, then add them.
Updated On: Jul 22, 2026
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The Correct Option is B
Solution and Explanation
Step 1: Set up the digit constraint.
Let the three digits of the number be \(a\), \(b\) and \(c\), each a whole number from 0 to 9 (and \(a\), the leading digit, at least 1 since it is a three digit number). We are told \(a \times b \times c = 70\).
Step 2: Break 70 into three digits.
Write 70 using its prime factors: \(70 = 2 \times 5 \times 7\). All three factors, 2, 5 and 7, are already single digits, so \(\{2, 5, 7\}\) is one valid set of digits with product 70.
Step 3: Check no other digit set gives 70.
For a different set of digits to multiply to 70, two of the prime factors would have to be combined into a single digit. Combining any two of 2, 5, 7 gives 10, 14 or 35, and each of these is bigger than 9, so it cannot be one digit. Bringing a 1 into the mix does not help either, since then the remaining two digits would need to multiply to 70, and no pair of digits from 1 to 9 does that (the biggest such product, \(9 \times 9\), is only 81, and checking every divisor pair of 70 shows none of them are both single digits). So 2, 5 and 7 are the only three digits whose product is 70.
Step 4: Add the digits.
Whatever order these digits appear in the number (257, 275, 527, 572, 725 or 752), the digit sum stays the same: \(2 + 5 + 7 = 14\).
Final Answer:
The sum of the digits is 14. \[ \boxed{14} \]
Let the three digits of the number be \(a\), \(b\) and \(c\), each a whole number from 0 to 9 (and \(a\), the leading digit, at least 1 since it is a three digit number). We are told \(a \times b \times c = 70\).
Step 2: Break 70 into three digits.
Write 70 using its prime factors: \(70 = 2 \times 5 \times 7\). All three factors, 2, 5 and 7, are already single digits, so \(\{2, 5, 7\}\) is one valid set of digits with product 70.
Step 3: Check no other digit set gives 70.
For a different set of digits to multiply to 70, two of the prime factors would have to be combined into a single digit. Combining any two of 2, 5, 7 gives 10, 14 or 35, and each of these is bigger than 9, so it cannot be one digit. Bringing a 1 into the mix does not help either, since then the remaining two digits would need to multiply to 70, and no pair of digits from 1 to 9 does that (the biggest such product, \(9 \times 9\), is only 81, and checking every divisor pair of 70 shows none of them are both single digits). So 2, 5 and 7 are the only three digits whose product is 70.
Step 4: Add the digits.
Whatever order these digits appear in the number (257, 275, 527, 572, 725 or 752), the digit sum stays the same: \(2 + 5 + 7 = 14\).
Final Answer:
The sum of the digits is 14. \[ \boxed{14} \]
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