SAT Math 2 problem/explanation

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1

Hi, I’m reading a McGraw-Hill book on the SAT Math 2 exam. There is a particular problem whose explanation I do not understand. Here’s the problem:

“The product of 45,454,545,454,545 and 1,234 contains how many digits? (A) 14, (B) 15, © 16, (D) 17, (E) 18”

It’s not too difficult a problem, but the explanation makes no sense to me. Here’s the explanation:

"Multiplying 45,454,545,454,545 by the units digit of 1,234, 4, results in a 14-digit product. Multiplying 45,454,545,454,545 by the tens digit of 1,234, 3, results in a 15-digit product because it is necessary to use a zero placeholder for the units digit. Similarly, multiplying by the hundreds digit requires 2 placeholders, and multiplying by the thousands digit requires 3 placeholders. The product will contain:

14 + 3 = 17 digits

On your calculator, the product may be displayed as 5.609 … E16, which represents 5.609 … × 10¹⁶. In decimal form, this results in a 5 followed by 16 digits."

I have a hard time following any of this explanation, but there’s one thing I’d like to single out. Namely, it’s just not true that multiplying that big number by 4 results in a 14-digit product. In fact, it results in a 15-digit product: 181,818,181,818,180.

Can anyone explain what this explanation is trying to say? Thanks!

2

@supplementfacts I have to agree with you - I don’t really understand their explanation either. In fact, 4545…45 x 30 is a 16-digit number, not 15, and the sum of a 17-digit number with a 16-digit number isn’t necessarily a 17-digit number, although it is here (for example, their solution fails at 88888888888888 x 1234) My best advice is to ignore that solution.

The problem is easily and correctly solved by noting that 4545…45 x 1000 has 17 digits, and 4545…45 x 2000 also has 17 digits (since it’s strictly less than 50,000,000,000,000 x 2000 = 10^17), so therefore 4545…45 x 1234 also has 17 digits.