Number of divisors and Integer | B.Stat Objective | TOMATO 83
Try this TOMATO problem from I.S.I. B.Stat Entrance Objective Problem based on Number of divisors and Integer.
Number of divisors and Integer (B.Stat Objective)
The smallest positive integer n with 24 divisors (where 1 and n are also considered divisors of n) is
Key Concepts
Check the Answer
Answer: 360
B.Stat Objective Question 83
Challenges and Thrills of Pre-College Mathematics by University Press
Try with Hints
number of divisors of 420=(2+1)(1+1)(1+1)(1+1)=24
number of divisors of 240=(4+1)(1+1)(1+1)=20
number of divisors of 360=(3+1)(2+1)(1+1)=24 and number of divisors of 480=(5+1)(1+1)(1+1)=24 then required number =360.
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Number of Factors | TOMATO B.Stat Objective 95
Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Number of factors.
Number of Factors (B.Stat Objective Question)
The number of different factors of 1800 equals
Key Concepts
Check the Answer
Answer: 36
B.Stat Objective Problem 95
Challenges and Thrills of Pre-College Mathematics by University Press
Try with Hints
here 1800=(2)(2)(2)(3)(3)(5)(5)
number of divisors of 1800 =(3+1)(2+1)(2+1)
=(4)(3)(3)=36.
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Number of divisors and Integers | TOMATO B.Stat Objective 97
Try this problem from I.S.I. B.Stat Entrance Objective Problem based on Number of divisors and Integers.
Number of divisors and Integers (B.Stat Objective Question)
The number of different factors of 6000, where 1 and 6000 are also considered as divisors of 6000, is
Key Concepts
Check the Answer
Answer: 40
B.Stat Objective Problem 97
Challenges and Thrills of Pre-College Mathematics by University Press
Try with Hints
Here 6000=(2)(2)(2)(2)(3)(5)(5)(5)
number of divisors of 6000 =(4+1)(1+1)(3+1) where number of divisors=(a+1)(b+1)(c+1) for n=\(p_1^{a}p_2^{b}p_3^{c}\) as \(p_1,p_2,p_3\) are primes
=(5)(2)(4)=40.
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Problem on Positive Integer | AIME I, 1995 | Question 6
Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1995 based on Problem on Positive Integer.
Problem on Positive Integer - AIME I, 1995
Let \(n=2^{31}3^{19}\),find number of positive integer divisors of \(n^{2}\) are less than n but do not divide n.
- is 107
- is 589
- is 840
- cannot be determined from the given information
Key Concepts
Check the Answer
Answer: is 589.
AIME I, 1995, Question 6
Elementary Number Theory by David Burton
Try with Hints
Let \(n=p_1^{k_1}p_2^{k_2}\) for some prime \(p_1,p_2\). The factors less than n of \(n^{2}\)
=\(\frac{(2k_1+1)(2k_2+1)-1}{2}\)=\(2k_1k_2+k_1+k_2\)
The number of factors of n less than n=\((k_1+1)(k_2+1)-1\)
=\(k_1k_2+k_1+k_2\)
Required number of factors =(\(2k_1k_2+k_1+k_2\))-(\(k_1k_2+k_1+k_2\))
=\(k_1k_2\)
=\(19 \times 31\)
=589.
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