Difference between revisions of "2006 iTest Problems/Problem 1"
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− | A [[divisor]] could include or exclude 2, include or exclude 17, and include or exclude 59. Thus, there are <math>2^3 = \boxed{\textbf{(A)} 8}</math> positive integral divisors. We can also note that answer choice A is the only answer choice and simply selected the option from the start. | + | A [[divisor]] could include or exclude 2, include or exclude 17, and include or exclude 59. Thus, there are <math>2^3 = \boxed{\textbf{(A) } 8}</math> positive integral divisors. We can also note that answer choice A is the only answer choice and simply selected the option from the start. |
==See Also== | ==See Also== | ||
− | + | {{iTest box|year=2006|before=First Problem|num-a=2|ver=[[2006 iTest Problems/Problem U1|U1]] '''•''' [[2006 iTest Problems/Problem U2|U2]] '''•''' [[2006 iTest Problems/Problem U3|U3]] '''•''' [[2006 iTest Problems/Problem U4|U4]] '''•''' [[2006 iTest Problems/Problem U5|U5]] '''•''' [[2006 iTest Problems/Problem U6|U6]] '''•''' [[2006 iTest Problems/Problem U7|U7]] '''•''' [[2006 iTest Problems/Problem U8|U8]] '''•''' [[2006 iTest Problems/Problem U9|U9]] '''•''' [[2006 iTest Problems/Problem U10|U10]]}} | |
− | |||
− | [[Category:Introductory | + | [[Category:Introductory Number Theory Problems]] |
Revision as of 01:15, 26 November 2018
Problem
Find the number of positive integral divisors of 2006.
Solution
First, factor the number 2006.
A divisor could include or exclude 2, include or exclude 17, and include or exclude 59. Thus, there are positive integral divisors. We can also note that answer choice A is the only answer choice and simply selected the option from the start.
See Also
2006 iTest (Problems, Answer Key) | ||
Preceded by: First Problem |
Followed by: Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • U1 • U2 • U3 • U4 • U5 • U6 • U7 • U8 • U9 • U10 |