Math Problems!........share It Here!

[st]

ano mag tatayo na ba tayo ng mtc math club?

Kayo po bahala

[PageDown]

A 3-digit number has its digit in arithmetic progression.

If 596 is added to this 3-digit number, the sum will have its digit in geometric progression.

What is the number?

Please post your solution.

[r3mu503]

246

 

I don't know enough of number theory, so this is what I used instead to get the answer (it's real clunky, so apologies ):

 

The largest sum possible for two 3-digit numbers is 1998, so _IF_ the sum of the mystery number and 596 results in a 4 digit number then this 4 digits would look like either of these two (since they are in geo. progression):

 

b b*r b*r^2 b*r^3

b*r^3 b*r^2 b*r b

 

(where each digit <=9)

 

With the leftmost digit (either 'b' or 'b*r^3') being a 1 (since we're checking the case where the sum is a 4 digit number).

 

So now we have either of these two conditions (since the leftmost digit is a 1):

 

b=1

b*r^3=1

 

In the first case, if b=1 then r=1 or 2 (since in this case, the rightmost digit 'b*r^3' must be <=9, thus r can at most only be equal to 2). Thus the possible 4-digit numbers here would be 1111 (b=1,r=1) and 1248 (b=1,r=2). Subtracting 596 from these yields 515 and 652, niether of which are in arith. progress. So this case is false.

 

If we take the second case, b*r^3=1, then b=r=1. Which again yields the possible 4-digit number 1111, subtracting 596 gives 515 which again isn't in arith. progress. so this case is false as well.

 

So the above shows that the sum of the mystery number and 596 cannot be a 4-digit number.

 

Ok, so now we know the sum is a 3-digit number, hence in the following form:

 

b b*r b*r^2

b*r^2 b*r b

 

We also know each digit <=9 and since 596 is being added the leftmost digit must >=6, thus:

 

9 >= b >= 6

9 >= b*r^2 >= 6

 

Taking the second case, b*r^2, there are only 3 possible values allowed for r (due to the <=9 condition for each digit), hence drawing a matrix for values of b given the allowed values for r:

 

b*r^2 b (r=1) b (r=2) b (r=3)

------- -------- --------- ---------

6 6 X X

7 7 X X

8 8 2 X

9 9 X 1

 

('X' marks non-valid combinations)

 

Which gives 6 possible 3-digit numbers for the sum of the mystery number and 596:

 

666 (b=6,r=1)

777 (b=7,r=1)

888 (b=8,r=1)

999 (b=9,r=1)

842 (b=2,r=2)

931 (b=1,r=3)

 

Subtracting 596 from each will show that only 842 will give you a 3-digit number (i.e. 246) which has it's digits in arith. progress.

 

Now for the first case, 9>=b>=6 ... nah, I'm too lazy, besides I've already got at least one answer

[Chinese Chicken]

246

 

I don't know enough of number theory, so this is what I used instead to get the answer (it's real clunky, so apologies ):

 

The largest sum possible for two 3-digit numbers is 1998, so _IF_ the sum of the mystery number and 596 results in a 4 digit number then this 4 digits would look like either of these two (since they are in geo. progression):

 

b b*r b*r^2 b*r^3

b*r^3 b*r^2 b*r b

 

(where each digit <=9)

 

With the leftmost digit (either 'b' or 'b*r^3') being a 1 (since we're checking the case where the sum is a 4 digit number).

 

So now we have either of these two conditions (since the leftmost digit is a 1):

 

b=1

b*r^3=1

 

In the first case, if b=1 then r=1 or 2 (since in this case, the rightmost digit 'b*r^3' must be <=9, thus r can at most only be equal to 2). Thus the possible 4-digit numbers here would be 1111 (b=1,r=1) and 1248 (b=1,r=2). Subtracting 596 from these yields 515 and 652, niether of which are in arith. progress. So this case is false.

 

If we take the second case, b*r^3=1, then b=r=1. Which again yields the possible 4-digit number 1111, subtracting 596 gives 515 which again isn't in arith. progress. so this case is false as well.

 

So the above shows that the sum of the mystery number and 596 cannot be a 4-digit number.

 

Ok, so now we know the sum is a 3-digit number, hence in the following form:

 

b b*r b*r^2

b*r^2 b*r b

 

We also know each digit <=9 and since 596 is being added the leftmost digit must >=6, thus:

 

9 >= b >= 6

9 >= b*r^2 >= 6

 

Taking the second case, b*r^2, there are only 3 possible values allowed for r (due to the <=9 condition for each digit), hence drawing a matrix for values of b given the allowed values for r:

 

b*r^2 b (r=1) b (r=2) b (r=3)

------- -------- --------- ---------

6 6 X X

7 7 X X

8 8 2 X

9 9 X 1

 

('X' marks non-valid combinations)

 

Which gives 6 possible 3-digit numbers for the sum of the mystery number and 596:

 

666 (b=6,r=1)

777 (b=7,r=1)

888 (b=8,r=1)

999 (b=9,r=1)

842 (b=2,r=2)

931 (b=1,r=3)

 

Subtracting 596 from each will show that only 842 will give you a 3-digit number (i.e. 246) which has it's digits in arith. progress.

 

Now for the first case, 9>=b>=6 ... nah, I'm too lazy, besides I've already got at least one answer

first paragraph pa lang nawala na ako.

[Guest -=xerxes=-]

246

 

I don't know enough of number theory, so this is what I used instead to get the answer (it's real clunky, so apologies ):

 

The largest sum possible for two 3-digit numbers is 1998, so _IF_ the sum of the mystery number and 596 results in a 4 digit number then this 4 digits would look like either of these two (since they are in geo. progression):

 

    b        b*r      b*r^2    b*r^3

b*r^3    b*r^2      b*r        b

 

(where each digit <=9)

 

With the leftmost digit (either 'b' or 'b*r^3') being a 1 (since we're checking the case where the sum is a 4 digit number).

 

So now we have either of these two conditions (since the leftmost digit is a 1):

 

b=1

b*r^3=1

 

In the first case, if b=1 then r=1 or 2  (since in this case, the rightmost digit 'b*r^3' must be <=9, thus r can at most only be equal to 2). Thus the possible 4-digit numbers here would be 1111 (b=1,r=1) and 1248 (b=1,r=2). Subtracting 596 from these yields 515 and 652, niether of which are in arith. progress. So this case is false.

 

If we take the second case, b*r^3=1, then b=r=1. Which again yields the possible 4-digit number 1111, subtracting 596 gives 515 which again isn't in arith. progress. so this case is false as well.

 

So the above shows that the sum of the mystery number and 596 cannot be a 4-digit number.

 

Ok, so now we know the sum is a 3-digit number, hence in the following form:

 

    b      b*r      b*r^2

b*r^2    b*r      b

 

We also know each digit <=9 and since 596 is being added the leftmost digit must >=6, thus:

 

9 >=    b    >= 6

9 >= b*r^2 >= 6

 

Taking the second case, b*r^2, there are only 3 possible values allowed for r (due to the <=9 condition for each digit), hence drawing a matrix for values of b given the allowed values for r:

 

b*r^2    b (r=1)    b (r=2)    b (r=3)

-------    --------    ---------    ---------

  6            6              X              X

  7            7              X              X

  8            8              2              X

  9            9              X              1

 

('X' marks non-valid combinations)

 

Which gives 6 possible 3-digit numbers for the sum of the mystery number and 596:

 

666 (b=6,r=1)

777 (b=7,r=1)

888 (b=8,r=1)

999 (b=9,r=1)

842 (b=2,r=2)

931 (b=1,r=3)

 

Subtracting 596 from each will show that only 842 will give you a 3-digit number (i.e. 246) which has it's digits in arith. progress.

 

Now for the first case, 9>=b>=6 ... nah, I'm too lazy, besides I've already got at least one answer

it can be solved in a much easier manner

Edited August 24, 2004 by -=xerxes=-

[oric91]

Write 1/2 as a decimal number without using the digit 5 nor use any signs of operator - plus, minus, and such in your solution. For example,

1/2= 0.6-0.1, is not considered a right solution.

[r3mu503]

it can be solved in a much easier manner

well dang man! don't just say it, post it here

 

I did say at the top of my post it was a clunky sol'n

Edited August 24, 2004 by -=xerxes=-

[goku]

please post your answers na para ma-post ko bagong puzzle about balls

[PageDown]

Sorry, you've got the correct answer but a BAD solution, as if you domn't know anything about math.

 

The answer is easy, but it's the way you solved it.

 

A don't see any progression formula.

 

When you say a 3-digit number, than it should be written as:

Let x = 100's digit, y = 10s didigt, and z = 1s digit

So the number would be 100x+10y+z = the number

 

This is how you solve a math problem... Ok?

[r3mu503]

Sorry, you've got the correct answer but a BAD solution, as if you domn't know anything about math.

 

The answer is easy, but it's the way you solved it.

 

A don't see any progression formula.

 

When you say a 3-digit number, than it should be written as:

Let x = 100's digit, y = 10s didigt, and z = 1s digit

So the number would be 100x+10y+z = the number

 

This is how you solve a math problem... Ok?

hi pagedown,

 

actually that's what I started with:

 

100x + 10(x+k) + (x+2k) + 596 = 100y + 10yc + yc*c

 

arith. progress. on the left (base of x, constant increment of k), geom. progress. on the right (base of y, constant multiplier of c)

 

but I wasn't getting anywhere using the above. I figured I lacked some know-how in number theory or sequences so I tried another approach. I didn't bother including the above anymore because it got me nowhere

 

and don't tell me I don't know math unless you know me, I have a BS in physics so I know enough math at least for that degree

[The_Blade]

Sorry, you've got the correct answer but a BAD solution, as if you domn't know anything about math.

 

The answer is easy, but it's the way you solved it.

 

A don't see any progression formula.

 

When you say a 3-digit number, than it should be written as:

Let x = 100's digit, y = 10s didigt, and z = 1s digit

So the number would be 100x+10y+z = the number

 

This is how you solve a math problem... Ok?

I think r3mu503 you presented the only answer. 246 + 596 = 842

 

solution no1. write all possible 3 digit #s which when you add to 596 will give you a 3 digit answer. the easiest solution, T&R. (trial and error).

 

123 596 719

135 596 731

147 596 743

159 596 755

234 596 830

246 596 842

258 596 854

321 596 917

345 596 941

357 596 953

369 596 965

 

solution no2. solve backwards. any 3 digit number when you add to 596 will always be greater than 596. now think of an integer larger than 5 but less than 9 (becoz the answer should be lesss than 999) which will give at least 3 factors. and could only be 8...4...2. or vice versa start with the last sensible digit. 1 is eliminated. 3 is disqualified becase it will give you 3..9...27 already. so 2 is the base number. 2..4..8. (process of elimination). now check your answer.

 

842-596=246, therefore 246 is the answer.

 

solution no.3 using algebra

 

Let x = 100's digit, y = 10s didigt, and z = 1s digit

So the number would be 100x+10y+z = N1(#in arithmitic progression)

 

Let 100a+10b+c =N2 (# in geometric progression)

where a=c^3; b=c^2 (now this is tricky, i assumed the 100digits is > 10 digit> 1 digit bcoz its the only sensible number > 596)

 

tf: 100c^3+10c^2+c=N2

 

100x+10y+z+596= 100c^3+10c^2+c ; 1 equation & 3 unknowns = indeterminate

 

but since 2 is the only sensible integer for c, by substitution:

 

100x+10y+z+ 596 = 100(8) + 10(4) + 2 = 842

 

100x+10y+z = 842-596 = 246 = 100(2)+10(4)+ 6

 

100x +10y + z = 100(2) + 10(4) + 6

 

therfore x=2 ; y=4 ; z=6

 

now boss xerxes and PageDown may i have your thoughts?

Edited July 30, 2004 by edzabu

[ukyo_batusai]

to dali lang.... find the product of....

 

 

( X - a ) ( X - b ) ( X - c )......( X - z )?

Edited July 30, 2004 by ukyo_batusai

[The_Blade]

to dali lang....  find the product of....

 

 

( X - a ) ( X - b ) ( X - c )......( X - z )?

X^26 +.................+ abc...z = 0 elementary!

 

 

the expression is also equal to:

 

(x-a) (x- (x-c) .........(x-x) (x-y) (x-z) but (x-x) = 0

 

therefore: any number multiplied by zero is equal to zero.

 

 

itaas ko lang meron pang tanong si oric91 na di nasasagot.

Edited August 2, 2004 by edzabu

[r3mu503]

hey, is the above post from ukyo a solution to a different question???

 

I still haven't seen the pure equation solution to the original question? you know, starting from:

 

100x + 10(x+k) + (x+2k) + 596 = 100y + 10yc + yc*c

 

and manipulating the above equation to prove the answer 246...

 

hi edzabu,

 

I think you'll need to use kc, kc^2, kc^3 for your geom. progression b/c that is strictly speaking the definition of a geom. progress. and the original poster did not say we could assume k=1 (hell, he actually didn't even say we could assume the order of the digits would be left to right in increasing value).

 

so where is the pure equation solution to this anyway???

[D_Engineer]

oppss..padaan lang po... Im an engineer by profession but mahina sa math...