Reasoning · Grade 5-1 Applications of Prime Factorization
Problem
Each trailing zero comes from one factor of 10.
Multiplying by 10 shifts every digit one place to the left and drops a zero into the ones place, so each 10 in a product buys exactly one zero at the end.
5.NBT.A.2Look For A PatternA 10 needs a 2 and a 5 together.
Pairing is the natural picture: think of 2s in one pile and 5s in another, and you can only make as many 10s as there are matching pairs, so the smaller pile is what limits you.
4.OA.B.4Identify SubproblemsEvery trailing zero comes from one 2 paired with one 5, so counting zeros means counting those pairs.
A number breaks into primes in only one way, so the tens hidden inside a product are exactly the twos and fives it contains.
One ten needs exactly one 2 and one 5, so tens and matched pairs of those primes correspond one for one.
(1) has five 2s and three 5s: 3 pairs.
Splitting each factor separately, then gathering like primes, keeps the bookkeeping simple and avoids multiplying anything out.
6.EE.A.1Make A Systematic ListSo (1) ends in 3 zeros.
The smaller count wins, and here the 5s are scarcer, which is the usual situation — 2s are easy to come by, 5s are not.
5.NBT.A.2Look For A Pattern(2) also makes three pairs: 3 zeros.
It is easy to miss that 35 hides a 5; splitting every factor into primes rather than eyeballing it is what catches the extra 5 and turns 2 zeros into 3.
6.EE.A.1Make A Systematic List(3) makes three pairs too: 3 zeros.
Note that 750 already ends in one zero, but that single zero is not the answer — the other factors supply two more 5s, and the pairing count is what settles it.
6.EE.A.1Make A Systematic ListAll three products end in 3 zeros.
Seeing the same answer three times is not a mistake — the products were built on purpose so that each carries exactly three 5s.
5.NBT.A.2Look For A PatternEvery zero at the end of a product comes from a 2 paired with a 5, so count the 2s, count the 5s, and the smaller count is your number of zeros.