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15 Apr 2026, 05:02
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
13% (01:31) correct
88%
(01:58)
wrong
based on 16
sessions
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For all positive integers x, f(x) is the number of positive divisors of x. If p is a positive integer less than 1000 such that f(2p) < f(3p) < f(5p) < f(7p) how many possible values of p are there?
A. 2
B. 3
C. 4
D. 5
E. more than 5
A. 2
B. 3
C. 4
D. 5
E. more than 5
15 Apr 2026, 14:28
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Answer is (A) 2.
This is a Number Properties question and the real work is figuring out what the condition f(2p) < f(3p) < f(5p) < f(7p) is actually saying about the prime factorization of p.
Key setup: For any prime q and positive integer p, f(qp) = f(p) * (v_q(p) + 2) / (v_q(p) + 1), where v_q(p) is the exponent of q in p's factorization. This ratio is 2 when q doesn't divide p, and decreases as the power of q in p increases.
So f(2p) < f(3p) means (v_2 + 2)/(v_2 + 1) < (v_3 + 2)/(v_3 + 1), which means v_2(p) > v_3(p). Similarly for the rest.
The condition simplifies to: v_2(p) > v_3(p) > v_5(p) > v_7(p) >= 0.
Since these are non-negative integers in strict descending order, the minimum possible values are v_7 = 0, v_5 = 1, v_3 = 2, v_2 = 3. So p must be divisible by at least 2^3 * 3^2 * 5 = 360.
Now I just enumerate what values of p under 1000 work:
1. p = 2^3 * 3^2 * 5 = 360: v_2=3, v_3=2, v_5=1, v_7=0. Check: 3>2>1>0. Valid.
2. p = 2^4 * 3^2 * 5 = 720: v_2=4, v_3=2, v_5=1, v_7=0. Check: 4>2>1>0. Valid.
3. Next option would be 2^3 * 3^2 * 5^2 = 1800 > 1000. Out.
4. Or 2^3 * 3^3 * 5 = 1080 > 1000. Out.
5. Adding any prime >= 11 to 360 gives 360*11 = 3960 > 1000. Out.
Verify with p = 360 = 2^3 * 3^2 * 5:
f(2*360) = f(2^4 * 3^2 * 5) = 5*3*2 = 30
f(3*360) = f(2^3 * 3^3 * 5) = 4*4*2 = 32
f(5*360) = f(2^3 * 3^2 * 5^2) = 4*3*3 = 36
f(7*360) = f(2^3 * 3^2 * 5 * 7) = 4*3*2*2 = 48
30 < 32 < 36 < 48. Confirmed.
So exactly 2 values of p work. The trap is assuming there are more solutions or jumping to "more than 5" -- the constraint is tight once you see what the inequality is really saying.
This is a Number Properties question and the real work is figuring out what the condition f(2p) < f(3p) < f(5p) < f(7p) is actually saying about the prime factorization of p.
Key setup: For any prime q and positive integer p, f(qp) = f(p) * (v_q(p) + 2) / (v_q(p) + 1), where v_q(p) is the exponent of q in p's factorization. This ratio is 2 when q doesn't divide p, and decreases as the power of q in p increases.
So f(2p) < f(3p) means (v_2 + 2)/(v_2 + 1) < (v_3 + 2)/(v_3 + 1), which means v_2(p) > v_3(p). Similarly for the rest.
The condition simplifies to: v_2(p) > v_3(p) > v_5(p) > v_7(p) >= 0.
Since these are non-negative integers in strict descending order, the minimum possible values are v_7 = 0, v_5 = 1, v_3 = 2, v_2 = 3. So p must be divisible by at least 2^3 * 3^2 * 5 = 360.
Now I just enumerate what values of p under 1000 work:
1. p = 2^3 * 3^2 * 5 = 360: v_2=3, v_3=2, v_5=1, v_7=0. Check: 3>2>1>0. Valid.
2. p = 2^4 * 3^2 * 5 = 720: v_2=4, v_3=2, v_5=1, v_7=0. Check: 4>2>1>0. Valid.
3. Next option would be 2^3 * 3^2 * 5^2 = 1800 > 1000. Out.
4. Or 2^3 * 3^3 * 5 = 1080 > 1000. Out.
5. Adding any prime >= 11 to 360 gives 360*11 = 3960 > 1000. Out.
Verify with p = 360 = 2^3 * 3^2 * 5:
f(2*360) = f(2^4 * 3^2 * 5) = 5*3*2 = 30
f(3*360) = f(2^3 * 3^3 * 5) = 4*4*2 = 32
f(5*360) = f(2^3 * 3^2 * 5^2) = 4*3*3 = 36
f(7*360) = f(2^3 * 3^2 * 5 * 7) = 4*3*2*2 = 48
30 < 32 < 36 < 48. Confirmed.
So exactly 2 values of p work. The trap is assuming there are more solutions or jumping to "more than 5" -- the constraint is tight once you see what the inequality is really saying.
