Question

Difficulty: HardExponents, Roots, and Powers of Integers

If nn is a positive integer greater than 2020 such that 228+220+2n\sqrt{2^{28} + 2^{20} + 2^n} is an integer, what is the least possible value of nn?

  1. A
    22
  2. B
    24
  3. 25Answer
  4. D
    26
  5. E
    34

Answer

The least possible value of nn is 25.
The expression inside the radical must be a perfect square. Matching 228+220+2n2^{28} + 2^{20} + 2^n to (214+210)2=228+2214210+220=228+225+220(2^{14} + 2^{10})^2 = 2^{28} + 2 \cdot 2^{14} \cdot 2^{10} + 2^{20} = 2^{28} + 2^{25} + 2^{20} reveals that 2n=2252^n = 2^{25}, so n=25n = 25. Since 25 is greater than 20 and smaller than the other valid solution n=34n = 34, it is the least possible value.

Step-by-Step Solution

1
Set up the perfect square structure for the radical expression.
For 228+220+2n\sqrt{2^{28} + 2^{20} + 2^n} to be an integer, the expression 228+220+2n2^{28} + 2^{20} + 2^n must equal (2a+2b)2=22a+2a+b+1+22b(2^a + 2^b)^2 = 2^{2a} + 2^{a+b+1} + 2^{2b} for some positive integers a>ba > b.
Expanding a binomial power of 2 generates three power-of-2 terms matching the three terms in the expression.
2
Analyze Case 1 where 2282^{28} corresponds to the leading term 22a2^{2a}.
2a=28    a=142a = 28 \implies a = 14. The remaining terms 2202^{20} and 2n2^n must correspond to 22b2^{2b} and 2a+b+12^{a+b+1}.
Matching highest powers establishes the value of the parameter aa.
3
Evaluate sub-cases for matching 2202^{20}.
Subcase 1: If 2b=20    b=102b = 20 \implies b = 10, then the middle term exponent is a+b+1=14+10+1=25a + b + 1 = 14 + 10 + 1 = 25. Thus n=25n = 25.
Subcase 2: If a+b+1=20    14+b+1=20    b=5a + b + 1 = 20 \implies 14 + b + 1 = 20 \implies b = 5, then 2b=102b = 10. Thus n=10n = 10.
Checking both term assignments determines possible integer values for nn.
4
Analyze Case 2 where 2202^{20} corresponds to 22b2^{2b} and 2282^{28} is the middle term.
If 2b=20    b=102b = 20 \implies b = 10 and a+b+1=28    a+11=28    a=17a + b + 1 = 28 \implies a + 11 = 28 \implies a = 17, then 2a=342a = 34. Thus n=34n = 34.
Checking alternative middle term assignments finds all valid values for nn.
5
Filter by constraint n>20n > 20 and select the minimum.
The valid values satisfying n>20n > 20 are n=25n = 25 and n=34n = 34. The least possible value is 25.
The question asks specifically for the least possible value greater than 20.

Key Concept

Perfect Square Trinomial Expansion with Exponent Rules
Estimated Time:2m 0s
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