A programme who differentiates 2 integer wether its odd or even

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Posted on Jan 02, 2017

Let x be the smallest number.

Let x + 2 be the other number (consecutive even integer)

Now to translate the rest;)

three times the smaller 3(x)

19 more -19

sum of the two integers - (x) +( x+2)

Pulling it together,

3x -19 = x + x +2

collect like terms

3x - 19 = 2x + 2

Put all the constants on one side by adding 19 to both sides.

3x - 19 +19= 2x + 2 + 19

3x = 2x +21

Subtract 2x from both sides to have all the x's on one side.

3x - 2x = 2x +21 - 2x

x = 21

The other number is 21 + 2, or 23

Check:three times the smaller = 3 x 21 = 63

sum of the two integers = 21 + 23 or 44

is 63 at least 19 more than 44

Let x + 2 be the other number (consecutive even integer)

Now to translate the rest;)

three times the smaller 3(x)

19 more -19

sum of the two integers - (x) +( x+2)

Pulling it together,

3x -19 = x + x +2

collect like terms

3x - 19 = 2x + 2

Put all the constants on one side by adding 19 to both sides.

3x - 19 +19= 2x + 2 + 19

3x = 2x +21

Subtract 2x from both sides to have all the x's on one side.

3x - 2x = 2x +21 - 2x

x = 21

The other number is 21 + 2, or 23

Check:three times the smaller = 3 x 21 = 63

sum of the two integers = 21 + 23 or 44

is 63 at least 19 more than 44

Feb 19, 2015 | Office Equipment & Supplies

Here's a simple (and inefficient) 35s program. It simply goes through a loop 25 times, adding up the numbers.

A001 LBL A

A002 25

A003 STO A

A004 0

A005 RCL+ A

A006 DSE A

A007 GTO A005

The sum of the first N positive integers can be calculated more efficiently as N(N+1)/2 .

A001 LBL A

A002 25

A003 STO A

A004 0

A005 RCL+ A

A006 DSE A

A007 GTO A005

The sum of the first N positive integers can be calculated more efficiently as N(N+1)/2 .

Mar 10, 2014 | HP 35s Programmable Scientific Calculator...

There is no solution in the integers. 20*d is an even number for all d, and no even number mod another even number is odd.

May 08, 2013 | Casio FX991ES Scientific Calculator

Let x be first (smaller) integer and y second (larger) integer.

These two numbers are 20 and 22.

- They are consecutive (and even) : y=x+2
- Twice the smaller exceeds the larger by 18: 2*x=y+18
- Substitute first equation into second: 2x=x+2+18
- 2x-x=2+18
- x=20
- y=x+2=22

These two numbers are 20 and 22.

Oct 03, 2011 | Office Equipment & Supplies

Let x be the smallest integer, then x+2 is the middle and x+4 the largest one. Resolve the equation

x + 2 * ( x + 4 ) = 5 * ( x + 2 ) - 24

=> x = 11

The three numbers are therefore 11, 13, and 15. The fact they are odd is irrelevant for finding the answer.

x + 2 * ( x + 4 ) = 5 * ( x + 2 ) - 24

=> x = 11

The three numbers are therefore 11, 13, and 15. The fact they are odd is irrelevant for finding the answer.

Jan 17, 2011 | Texas Instruments TI-84 Plus Calculator

X + ( X + 2 ) + ( X + 4 ) = 105

3X + 6 = 105

3X = 105 - 6

3X = 99

X = 99/3

X = 33

So 33 + 35 + 37 = 105

3X + 6 = 105

3X = 105 - 6

3X = 99

X = 99/3

X = 33

So 33 + 35 + 37 = 105

Oct 06, 2010 | American Standard Plumbing

2n = 1st even integer

2n + 2 = 2nd even integer

Product = 2n(2n + 2) = 4n^2 + 4n = 4(n^2 + n)

1520 = 4(n^2 + n)

1520/4 = 4(n^2 + n)/4

380 = n^2 + n

0 = n^2 + n - 380

0 = (n + 20)(n - 19)

n = -20, +19

Answer can not be negative so n = +19

2n = 2(19) = 38

2n + 2 = 2(19) + 2 = 40

ANSWER: the integers are 38 and 40

2n + 2 = 2nd even integer

Product = 2n(2n + 2) = 4n^2 + 4n = 4(n^2 + n)

1520 = 4(n^2 + n)

1520/4 = 4(n^2 + n)/4

380 = n^2 + n

0 = n^2 + n - 380

0 = (n + 20)(n - 19)

n = -20, +19

Answer can not be negative so n = +19

2n = 2(19) = 38

2n + 2 = 2(19) + 2 = 40

ANSWER: the integers are 38 and 40

Nov 16, 2009 | RCA HD52W59 HDTV Projection Television

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