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Find the sum of all integers between 84 and 716 which are divisible by 5.

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Posted on Sep 15, 2011

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Search a string of at least five numbers (for example, 65403) Identify all of the substrings that form numbers that are divisible by 3.


I guess a hint is that to see if a number is divisible by 3, you can add up the digits of the number and if the sum is divisible by 3, the number is divisible by 3. Another way of seeing if a number is divisible by 3 is to do an integer divide and then multiply back to see if you get what you started with. For example, 7 / 3 = 2. 2 * 3 = 6, thus 7 is not divisible by 3.

As for all the substrings, start at the 5th number and do substrings of length 1 to 5, Move to the next digit, do substrings of length 1 to 4. Move to the next digit, do substrings of length 1 to 3. Repeat until only 1 digit is left.

Good luck,

Paul

Apr 12, 2016 | Office Equipment & Supplies

2 Answers

The numbers on three raffle tickets are consecutive integers whose sum is 7530. find the integers


The integers of the numbers on three raffle tickets are consecutive integers whose sum is 7,530 are 2509, 2510, and 2511.

Feb 24, 2015 | SoftMath Algebrator - Algebra Homework...

1 Answer

Is 450 divisible by 3 in fractions


450=9*5*10=9*50
9*50/3=3*50=150
To verify if a number is divisible by 3, add the digits in the number. If the sum of digits is divisible by 3, then thumber is also divisible by 3.
Sum of digits in 450 =4+5+0 =9, divisible by 3, so 450 is also divisible by 3.

Jul 06, 2014 | Office Equipment & Supplies

1 Answer

How do i factor numbers on a ti-30xs calculator


This calculator does not have a key that you can use to find the prime factor of an integer.
You can however use the calculator to find the factors
1. If number is even divide it by 2
Keep dividing by 2, while keeping track of how many times you divided by 2.
If you divided 5 times by 2 before getting an odd number, then your first factor is 2^5
2. Now try dividing by 3, keep track of the number of times you divided by 3 before you could not divide by 3 any more. If you divided 0 times by 3, your second factor is 3^0, or 1 and 3 is not a factor.
If you divided 4 times by 3, your second factor is 3^4
3. Divide by 5, until you can't any more. Keep track of the number of times you divided by 5.
4. Divide by all other prime numbers 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, etc. For each prime number, keep track of how many times you divided by it, until you could not any more.

Example: 23100
Division by 2
23100/2=11550 ---------> 1 division by 2
11550/2=5775 ----------> 2 divisions by 2
Note that 5775 is not divisible by 2. No more divisions by 2.
First factor is 2^2


Division by 3:
5775/3=1925 ----------> 1 division by 3
1925/3=641. 66667 Not an integer. No more divisions by 3.
2nd factor is 3^1

Division by 5 (number ends in 5)
1925/5 =385 -------> 1 division by 5
385/5=77 ----------> 2 divisions by 5 and no more (quotient does not end in 0 or 5)
3rd factor is 5^2

Division by 7
77/7=11 --------> 1 division by 7, and no more
4th factor is 7^1=7

Division by 11
11/11=1
5th factor is 11^1=11

Assembling the factors 2^2, 3^1, 5^2, 7, 11
Prime factorization of 23100 is
23100=(2^2)(3)(5^2)(7)11

Dec 18, 2013 | Texas Instruments TI-30XA Calculator

1 Answer

How to calculate 5^(-1)mod 792


I am sorry, but I was under the impression that the modulo operation involves integer divisions only. This means it involves only integers. However 5^(-1) is a rational number that is not an integer. And then I could be wrong.
As to doing modulo operations with this calculator, well you can't. Sorry again.

Nov 19, 2012 | Casio FX82MS Scientific Calculator

1 Answer

Ive.comprogram to find the number of and sum of all integers greater than 40 and less than 250 that are divisible by 5 in java coding ????


I'm not going to do your homework for you, but here's a hint. You can use the Java modulus operator % to find if a number is divisible by 5. It provides the remainder after dividing, which will be zero if the number is divisible by 5. Here's a pseudocode example:

if ( (testNumber % 5) == 0)

{
numberIsDivisibleBy5 = true
}

else

{
numberIsDivisibleBy5 = false
}

You'll need to provide the rest of the work to put this test into a loop that goes from >40 to <250, checking each number and doing whatever you are supposed to (print a message, probably) when you find one divisible by 5.

Good luck and thanks for using Fixya.

Jan 26, 2011 | Computers & Internet

1 Answer

Write a program to sum of all even integer numbers


#include <iostream>
using namespace std;
int main ()
{
int num = 0, i = 1, sum = 0;
cout << "Enter a positive integer: ";
cin >> num;
while (i <= num)
{
sum += i;
i++;
}
cout << "The sum is " << sum << endl;
return 0;
}

Oct 30, 2010 | Compaq Visual Fortran Professional Edition...

1 Answer

What are the three consecutive integers whose sum is 378?


Let X be the lowest of the 3 integers
The sum of the 3 integers can be represented as X + (X+1) + (X+2).
Set this sum equal to 378 and solve for X.
X + (X+1) + (X+2) = 378
3X + 3 = 378
3X = 375
X = 125
Since X is the lowest of the 3 integers the other 2 will be X+1 and X+2 or 126 and 127.

Therefore, the answer is 125, 126 and 127

Aug 23, 2010 | SoftMath Algebrator - Algebra Homework...

1 Answer

Pls. help me


Q.2#  Dim n, sum as Double
Dim i as Integer
n=100
sum=0

for i=1 to n
sum=sum+1/n
next i

Q.3#  Dim sum as Double
sum=0
for i=1 to 99
if i%2 <> 0 then sum=sum+i end if
next i

Sep 16, 2008 | Microsoft AccRepair DS2 2004 for Fr...

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