Need help in these problems

[bluecoal]

There are numerous problems I need some help:

1. Let the positive integer n>=2. A, B and C are distinct integers base 10, in such a way the Integer A is formed by n integer digits x while B is formed by identical digits y and C is formed by 2n identical digits z. Find all possible values odds x,y and z so that A² + B = C.

2. Let abc be a three-digit number such that the sum of acb, bca, bac, cab,cba is N if N = 3194, then what is three digit number abc?

3. Find all possible 4-digit number abcd such that the sum of a 3-digit number abc and 246 is 111 times of the digit d. abc is a multiple of 18.

 

[jsterkel]

It is usually better to post only a single problem in a thread. Also, please show your work, so that we can know where to help you.

 

[jsterkel]

2. Let abc be a three-digit number such that the sum of acb, bca, bac, cab,cba is N if N = 3194, then what is three digit number abc?

100a + 200b + 200c + 20a + 10b + 10c + 2a + 2b + c = 3194

122a + 212b + 221c = 3194

Can you solve it from here?

 

[jsterkel]

3. Find all possible 4-digit number abcd such that the sum of a 3-digit number abc and 246 is 111 times of the digit d. abc is a multiple of 18.

abc + 246 = 111d

Let x be a number such that 18x = abc

18x + 246 = 111d

Divide both sides by 3, leaving
6x + 82 = 37d

We also know that d is a single digit, so we have at most 10 values (0, 37, 74, 111, 148, 185, 222, 259, 296, 333).

Can you solve for all values of x from here? Then multiply each value of x by 18 for abc.


Edited: the number "10" is not a single digit, so 370 is not a valid answer....

 

[burakaltr]

There are numerous problems I need some help:

1. Let the positive integer n>=2. A, B and C are distinct integers base 10, in such a way the Integer A is formed by n integer digits x while B is formed by identical digits y and C is formed by 2n identical digits z. Find all possible values odds x,y and z so that A² + B = C.

2. Let abc be a three-digit number such that the sum of acb, bca, bac, cab,cba is N if N = 3194, then what is three digit number abc?

3. Find all possible 4-digit number abcd such that the sum of a 3-digit number abc and 246 is 111 times of the digit d. abc is a multiple of 18.

1. n>=2

A=(aaaa..) n digits =x
B=(bbbbb...)n digits=y
C(ccccc...)2n digits=z

(....+100a+10a+a)**2 + (...+111b)=(...100000c+10000c+1000c+100c+10c+c)

((111111..)ntimes**2) *( a**2)+(1111..)ntimes*b=(11111...)2n times*c

(11111...)ntimes*[ (1111)ntimes*a**2+b)=(11111..)2n timesc= [(1111..000..)ntimeszero,ntimes1+(00001111..)ntimeszero,ntimes1]*c ]c

(1111...)ntimes *( a**2 +b)= [(111..)ntimes*10**n + (111..)ntimes]*c

at n=3

111(a**2 +b)=111*1000+111c

111(a**2+b-c)=111000

100(a**2+b-c)+10(a**2+b-c)+(a**2+b-c)=111000

1000=a**2+b-c

at n=4
1111(a**2 +b)=1111*1000+1111c

a**2+b-c=10000

SO generally

a**2 +b-c=10**n {Solve for a,b,c all odds where n>=2 }

for n=2 for example

a**2+b-c=100



what I make out is smt. like the above expression if that helps.

 

[soroban]

Jello, bluecoal!

A variation of jsterkel's solution . . .

3. Find all possible 4-digit number abcd such that the sum of the 3-digit number abc
and 246 is 111 times the digit d. .abc is a multiple of 18.

We are told: .\(\displaystyle {\bf abc} \,=\,18n\) .[1]

. . and that: .\(\displaystyle 18n + 246 \:=\:111d \quad\Rightarrow\quad n \:=\:\dfrac{111d-246}{18} \)

So we have: .\(\displaystyle n \:=\:6d - 13 + \dfrac{d-4}{6}\) .[2]

Since \(\displaystyle n\) is an integer, \(\displaystyle d-4\) must be divisible by 6.

Since \(\displaystyle d\) is a digit: .\(\displaystyle {\bf d} = 4\)

Substitute into [2]: .\(\displaystyle n \:=\:6(4) - 13 + \frac{4-4}{6} \quad\Rightarrow\quad n \:=\:11\)

Substitute into [1]: .\(\displaystyle {\bf abc} \:=\:18(11) \:=\:198\)


Therefore: .\(\displaystyle {\bf abcd} \;=\;1984\)

 

[soroban]

Hello again, bluecoal!

I have a start on the first one . . . if I interpret it correctly.

1. Let the integer \(\displaystyle n \ge 2.\)
\(\displaystyle A, B\) and \(\displaystyle C\) are distinct integers, base 10.
The integer \(\displaystyle A\) is formed by \(\displaystyle n\) digits \(\displaystyle x.\)
The integer \(\displaystyle B\) is formed by \(\displaystyle n\) digits \(\displaystyle y.\)
The integer \(\displaystyle C\) is formed by \(\displaystyle 2n\) digits \(\displaystyle z.\)

Find all possible values of odd digits \(\displaystyle x,y\) and \(\displaystyle z\) so that: .\(\displaystyle A^2 + B \:=\:C.\)

\(\displaystyle A \;=\;\underbrace{xxx\hdots x}_{n\text{ digits}} \;=\; \left(\dfrac{10^n-1}{9}\right)x \)

\(\displaystyle B \;=\;\underbrace{yyy\hdots y}_{n\text{ digits}} \;=\;\left(\dfrac{10^n-1}{9}\right)y \)

\(\displaystyle C \;=\;\underbrace{zzz\hdots z}_{2n\text{ digits}} \;=\;\left(\dfrac{10^{2n}-1}{9}\right)z\)


\(\displaystyle A^2 +B\,=\,C\)

. . \(\displaystyle \left(\dfrac{10^n-1}{9}\,x\right)^2 + \dfrac{10^n-1}{9}\,y \;=\;\dfrac{10^{2n}-1}{9}\,z\)

. . \(\displaystyle \left(\dfrac{10^n-1}{9}\right)^2x^2 + \left(\dfrac{10^n-1}{9}\right)y \;=\;\dfrac{(10^n-1)(10^n+1)}{9}z \)


Divide by \(\displaystyle \frac{10^n-1}{9}\!:\;\;\left(\dfrac{10^n-1}{9}\right)x^2 + y \;=\;(10^n+1)z\)

Multiply by 9: . \(\displaystyle (10^n-1)x^2 + 9y \;=\;9(10^n+1)z\)

Solve for \(\displaystyle y\!:\;\;y \;=\;(10^n+1)z - \dfrac{10^n-1}{9}x^2\)


Then \(\displaystyle x,y,z\) are odd digits which satisfy:

. . \(\displaystyle y \;=\; \underbrace{(100\hdots1)}_{n+1\text{ digits}} z - \underbrace{(111\hdots1)}_{n\text{ digits}} x^2 \)

Does this help?


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


By inspection, I see that: .\(\displaystyle z=1,\;x=3\) produces \(\displaystyle y = 2\;\; (even!)\)

It makes me suspect that there are no solutions.

 

[Mario]

5x(2+1=5