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airh3ad

ALGEBRA PRACTICEï¿½

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A man wanted to get into his work building, but he had forgotten his code.
However, he did remember five clues. These are what those clues were:

The fifth number plus the third number equals fourteen.

The fourth number is one more than the second number.

The first number is twice the second number less than one.

The second number plus the third number equals ten.

The sum of all five numbers is 30.

What were the five numbers and in what order?

endless

is it

7 - 4 - 6 - 5 - 8

airh3ad

You got it endless hehehe good job man..congrats! A7 - B4 -C 6 - D5 - E8 but i need also the solotion additional 5frih$ points.

endless

The fifth number plus the third number equals fourteen.

The fourth number is one more than the second number.

The first number is twice the second number less than one.

The second number plus the third number equals ten.

The sum of all five numbers is 30.

What were the five numbers and in what order?

how i found it:

in the first hint it says that it no 3 and no 5 equals 14
the last hint says that the sum of all number is 30
so i know that no 1 2 and 4 are 16 together
with the 2nd and 3th hint i just experimented with some digitys
with 4 you get:
no 1 = 7
no 2 = 4
no 4 = 5
all together: 16
the 4th hint says 2nd and 3th equal 10 so no 3 is 6
3 and 5 equal 16 so i know 5 is 8

is that what you meant with solution?

hunnyhiteshseth

Well I do not think thats the correct answer. What he gave was one of the solution but there exist more than one solution to this problem. Let me elaborate:

Let the 5 numbers be:-
a,b,c,d,e in order.

So, according to given statements:-

c+e=14----------------------------------(i)
d=b+1-----------------------------------(ii)
a=2b-1----------------------------------(iii)
a+b+c+d+e=30------------------------(iv)

Putting value of (i),(ii) & (iii) in (iv)

We get 4b=16
=> b=4

=> d=5

=> a=7

Now, there can be multiple pairs of c and e

Since c + e = 14 and c and e are single digit numbers. So, there can be following pairs possible:-

(9,5), (5,9)
(8,6), (6,8 )
(7,7)

endless

airh3ad

Here is my own solution

Let a = 1st number
b = 2nd number
c = 3rd number
d = 4th number
e = 5th number

from ist clue :

e + c + 14 eq. 1

from 2nd clue

d = 1 + b eq. 2

from 3rd clue

a + 2b - 1 eq.3

from 4rt clue

b + c = 10 eq. 4

from 5th clue

a+b+c+d+e = 30 eq.5

from eq. 1

e+c = 14

e=14-c eq.6

from eq.4

b+c = 10

c=10-b eq.7

e = 14 -c

e=14 (10-b)

e= 4 + b eq.8

By Subt.

from eq 5

a+b+c+d+e = 30

(2b-1)+b+(10-b)+(1+b)+(4+b)=30

2b-1+b+10-b+1+b+4+b=30

2b+b-b+b+b=30-4-1-10+1

4b= 16
b=4

from eq. 3 b=4
a=2b-1
a=2(4)-1
a=7

from eq.7 b=4
c=10-b
c=10-4
c=6

from eq.2 ; b=4

d= 1 +b
d=1+4
d=5

from eq.8: b=4
e=4+b
e=4+4
e=8

Combination code

A B C D E
1 4 6 5 8

thats great and detailed solotion of mine..
Razz

endless

thats the problem with people who studied algebra, they do it too detailed, keep it simple Razz

~
(maybe it's because im 14 that i think simple but Razz

)

hunnyhiteshseth

airh3ad

thats part of curriculum i think algebra is the subject in schools.