16 items are filed under this topic.
 |
|
|
|
|
|
|
|
|
Divisibility of 3n^5+7n |
2016-12-14 |
|
From Parag:
Question from Parag, a student:
if n is a natural number,then 3n^5+7n is divisible by
a)2
b)3
c)5
d)7
i got the answer but still i need a valid alternate approach.
Answered by Penny Nom. |
|
|
|
|
|
|
4821x14y is an 8-digit number divisible by 72 |
2014-08-06 |
|
From RAYA:
if 4821x14y is an 8-digit number divisible by 72. How many values can x and y take?
Answered by Penny Nom. |
|
|
|
|
|
|
The square of any odd number, decreased by 1, is divisible by 8 |
2012-11-16 |
|
From bailey:
Prove that the square of any odd number, decreased by 1, is divisible by 8
Answered by Penny Nom. |
|
|
|
|
|
|
The difference of the two numbers |
2010-02-15 |
|
From Steve:
The difference of the two numbers 'abcdef ' and ' fdebca ' is divisible by 271. prove
that b = d and c = e.
Answered by Claude Tardif. |
|
|
|
|
|
|
How many combinations of 8614 are divisible by 7? |
2008-01-22 |
|
From Rebecca:
How many combinations of 8614 are divisible by 7 equally (with no remainder)?
Answered by Penny Nom. |
|
|
|
|
|
|
Induction - divisibility |
2007-08-04 |
|
From Jerry:
How would you prove that for any positive integer n, the value of the expression 3^(2n+2) - 8n -9 is divisible by 64.
Answered by Chris Fisher and Penny Nom. |
|
|
|
|
|
|
Divisibility |
2007-05-18 |
|
From Ashish:
A number is divisible by 2^n if the last n digits of the number are divisible by 2^n.
Why?
Answered by Penny Nom and Claude Tardif. |
|
|
|
|
|
|
Divisibility by each of the first ten counting numbers |
2005-10-17 |
|
From Simon:
determine smallest positive integer that is divisible by each of the first ten counting numbers
Answered by Penny Nom. |
|
|
|
|
|
|
A 3 digit number divisible by 7 |
2004-05-03 |
|
From A student:
We need to arrange 1,3 and 6 to form a 3 digit number that is divisible by 7.
Answered by Penny Nom. |
|
|
|
|
|
|
Divisibility by 2 or 5 or both |
2003-10-30 |
|
From Abdu:
How many positive integers less than 1,001 are divisible by either 2 or 5 or both?
Answered by Penny Nom. |
|
|
|
|
|
|
Three consecutive positive intergers |
2003-02-09 |
|
From Yew:
Prove that when we multiply any consecutive positive intergers, the result is always divisible by 6.
ex. (7)(8)(9) = 504 = 6 (84)
Answered by Penny Nom. |
|
|
|
|
|
|
Divisibility of 5 2002 |
2002-08-25 |
|
From Simon:
I need to ask you a question if 5 2002 and 3 2002 are divisible by 26.
Answered by Penny Nom. |
|
|
|
|
|
|
The number of hidden cubes |
2002-02-05 |
|
From Katie:
This problem is about finding the number of cubes visible and hidden in a cube.
In a cube that is 3x3, 19 cubes can be seen. 8 are hidden.
In a cube that is 4x4, 37 cubes can be seen. 27 are hidden.
In a cube that is 5x5, 61 cubes can be seen. 64 are hidden.
In a cube that is 6x6, 91 cubes can be seen. 125 are hidden. The question is:
Explain how you could find the number of small cubes that are visible and hidden in a cube of any size.
Answered by Paul Betts and Penny Nom. |
|
|
|
|
|
|
Divisibility by 9 |
2000-10-24 |
|
From Kelera:
If the sum of the digits of a number is divisible by 9, then the number itself it divisible by 9. Why is that? How do you explain this?
Answered by Penny Nom. |
|
|
|
|
|
|
111...1222...2 |
1999-08-11 |
|
From Brad Goorman:
Let N = 111...1222...2, where there are 1999 digits of 1 followed by 1999 digits of 2.
Express N as the product of four integers, each of them greater than 1.
Answered by Penny Nom. |
|
|
|
|
|
|
Divisibility of 2n choose n. |
1996-09-24 |
|
From Kathy Doan:
Can you prove that "2n choose n" is not divisible by 3, 5, and 7 for infinitely many n?
Answered by Penny Nom. |
|
|
