Try it yourself first — the explanation is most useful after you’ve attempted it.
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Toolkit + CCSS Solution
Understand
Restated: Find every prime number that divides $2010$, then add those primes together.
Givens: The target number is $2010$; We want only the distinct prime factors — each prime is counted once even if it appears more than once; Answer choices: (A) $67$, (B) $75$, (C) $77$, (D) $201$, (E) $210$
Unknowns: The sum of the distinct prime factors of $2010$
Understand
Restated: Find every prime number that divides $2010$, then add those primes together.
Givens: The target number is $2010$; We want only the distinct prime factors — each prime is counted once even if it appears more than once; Answer choices: (A) $67$, (B) $75$, (C) $77$, (D) $201$, (E) $210$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #3 Eliminate Possibilities
Factoring $2010$ is too big to do in one shot, so Tool #7 (Identify Subproblems) breaks it into a chain of smaller divisions: peel off one small prime at a time ($2$, then $3$, then $5$, …) until what is left is itself prime. Each step is a tiny problem we can do mentally. Tool #3 (Eliminate Possibilities) is the multiple-choice safety net — once we have a candidate sum, we can check it against the listed choices and rule the others out.
Execute — Answer: C
#7 Identify Subproblems 4.OA.B.4Step 1
Peel off the smallest prime.
$2010$ ends in $0$, so it is even — divide by $2$.
$$2010 \div 2 = 1005$$
💡 Splitting off one prime at a time turns a big factoring task into a sequence of easy divisions — the Tool #7 move.
#7 Identify Subproblems 4.OA.B.4Step 2
Check $1005$ for the next small prime.
The digits sum to $1+0+0+5 = 6$, which is divisible by $3$, so $1005$ is divisible by $3$.
$$1005 \div 3 = 335$$
💡 The digit-sum divisibility rule is the fastest way to test for $3$ without long division.
#7 Identify Subproblems 4.OA.B.4Step 3
Check $335$ for divisibility by $5$.
It ends in $5$, so yes.
$$335 \div 5 = 67$$
💡 Numbers ending in $0$ or $5$ are always divisible by $5$.
#7 Identify Subproblems 4.OA.B.4Step 4
Decide whether $67$ is prime.
Test small primes up to $\sqrt{67} \approx 8.2$: $67$ is not even, digit sum $6+7=13$ is not a multiple of $3$, it doesn't end in $0$ or $5$, and $67 \div 7 = 9$ remainder $4$.
So $67$ is prime, and the factorization is complete.
$$2010 = 2 \times 3 \times 5 \times 67$$
💡 Once the leftover quotient is prime, the chain of subproblems is done.
#3 Eliminate Possibilities 4.NBT.B.4Step 5
Add the distinct prime factors to get the requested sum.
💡 $77$ matches choice (C); the other choices are quickly ruled out — (A) $67$ forgets to add $2+3+5$, (E) $210$ is just the four digits of $2010$ rearranged, and (D) $201$ is $2010 \div 10$, a distractor.
[1]
#7 4.OA.B.4Peel off the smallest prime. $2010$ ends in $0$, so it is even — divide by $2$.
[2]
#7 4.OA.B.4Check $1005$ for the next small prime. The digits sum to $1+0+0+5 = 6$, which is
[3]
#7 4.OA.B.4Check $335$ for divisibility by $5$. It ends in $5$, so yes.
[4]
#7 4.OA.B.4Decide whether $67$ is prime. Test small primes up to $\sqrt{67} \approx 8.2$: $
[5]
#3 4.NBT.B.4Add the distinct prime factors to get the requested sum.
Review
Reasonableness: Multiply the factorization back: $2 \times 3 = 6$, $6 \times 5 = 30$, $30 \times 67 = 2010$. The product checks, so the prime list $\{2, 3, 5, 67\}$ is correct, and $2 + 3 + 5 + 67 = 77$ is the right sum. The answer is in the same order of magnitude as the small primes plus one mid-sized prime, which matches the choice $(C) 77$.
Alternative: Tool #3 (Eliminate Possibilities) on its own: the four smallest distinct primes are $2, 3, 5, 7$, which sum to $17$. Each answer choice minus $2+3+5 = 10$ would have to be a prime factor of $2010$: (A) $67-10=57$ — not a factor; (B) $75-10=65$ — not prime; (C) $77-10=67$, and $2010/67 = 30 = 2 \times 3 \times 5$ — works; (D) and (E) give factors larger than $2010$. Only (C) survives.
CCSS standards used (min grade 4)
4.OA.B.4 Find factor pairs, recognize multiples, and identify prime numbers within $1$-$100$ (Peeling $2010$ apart by repeatedly dividing by small primes ($2$, $3$, $5$) and recognizing that the final quotient $67$ is itself prime — the Grade 4 prime-factorization standard.)
4.NBT.B.4 Fluently add and subtract multi-digit whole numbers using the standard algorithm (Adding the four distinct prime factors $2 + 3 + 5 + 67 = 77$ at the end.)
⭐ Big numbers like $2010$ become easy once you peel off small primes one at a time — a Grade 4 factor-finding skill is all you need!
⭐ Big numbers like $2010$ become easy once you peel off small primes one at a time — a Grade 4 factor-finding skill is all you need!
