Please help find double-digit number which is 19 times greater than 2nd digit

Martynas

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Feb 6, 2018
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The task is to find a double digit number which is 19 times greater than it's second digit
I made an equation which looked like this: 10a+b=19b (a is the first digit and b is the second one)
but there are two unknown values(a,b) so i need another equation but how would it look like because i cant think of any way i can make it.
The answer is easy to figure out because its just a double digit but how would i prove that i am not guessing. The answer is 95.
 

tkhunny

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The task is to find a double digit number which is 19 times greater than it's second digit
I made an equation which looked like this: 10a+b=19b (a is the first digit and b is the second one)
but there are two unknown values(a,b) so i need another equation but how would it look like because i cant think of any way i can make it.
The answer is easy to figure out because its just a double digit but how would i prove that i am not guessing. The answer is 95.
Another equation would be helpful, but it is not necessary. Let's see what else you know about these 'a' and 'b'?

10a + b = 19b -- That's where you started.

We're in a Base 10 Positional System, so both 'a' and 'b' must be from {0,1,2,3,4,...9}

Neither can be zero (0), since a = 0 makes a 1-digit number and b = 0 doesn't leave much for 19*b. Wait! Is 10 * 0 + 0 = 19*0? Interesting idea that we can keep in the back of our head.

b CANNOT be in {6, 7, 8, 9}, since 19 multiplied by any of these would result in a 3-digit number for 10a + b.

That just leaves us with b in {1, 2, 3, 4, 5}

Also, when multiplying 'b' by 9 (the one's digit of 19), the digit must remain the same.

1*9 = 9 -- Nope -- 1 is not 9
2*9 = 18 -- Nope -- 2 is not 8
3*9 = 27 -- Nope -- 3 is not 7
4*9 = 36 -- Nope -- 4 is not 6
I see a pattern!
5*9 = 45 -- We could have something!

Thinking. Hunting. Eliminating. Thinking. Reason it out.
 

Dr.Peterson

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Nov 12, 2017
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The task is to find a double digit number which is 19 times greater than it's second digit
I made an equation which looked like this: 10a+b=19b (a is the first digit and b is the second one)
but there are two unknown values(a,b) so i need another equation but how would it look like because i cant think of any way i can make it.
The answer is easy to figure out because its just a double digit but how would i prove that i am not guessing. The answer is 95.
Hi, Martynas.

If this were an ordinary algebra problem, then you would expect a second equation. But it is not; it is a Diophantine equation -- that is, one restricted to integer (or sometimes, rational) solutions. In a sense, this restriction (that a and b must both be whole numbers less than 10) is the "second fact" that allows it to have only one solution.

Commonly, guessing (that is, checking the possible solutions) is the easiest way, and perhaps the only way) to complete the solution of this sort of problem. There are a variety of techniques that can be used in more complicated cases, but it may still come down to listing and trying possibilities in the end.

Here is how I would approach it:

10a+b=19b implies 10a = 18b, which implies 5a = 9b.

Therefore b must be a multiple of 5 (so that both sides can be multiples of 5), and can only be either 0 or 5. We know 0 won't work; so b = 5 and a = 9. And we're done. (We could also have said that a must be a multiple of 9.)

So the first bit was algebra, but the rest was number theory (primes and divisibility). That's typical. And you can see that it was possible that we would still have more than one possibility in a different problem. (Sometimes it will be inequalities that rescue us in the end; for example, that a < 10.)
 

stapel

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The task is to find a double digit number which is 19 times greater than it's second digit
I made an equation which looked like this: 10a+b=19b (a is the first digit and b is the second one)
This equation says that the second digit, b, is nineteen times as much as (nineteen times) the whole two-digit number. The exercise says that b is nineteen times "greater than" (+19*) that digit. So, if we are to take the exercise literally, the equation should be as follows:

. . . . .10a + b = b + 19b

...which simplifies as:

. . . . .10a + b = 20b

...and then can be rearranged as:

. . . . .10a = 19b

. . . . .a = 1.9b

Then, since there are only nine values (1, 2,..., 8, 9) that can be plugged in for "a", you plug them in and see what you get. From the results, you'll quickly see that the exercise was either copied incorrectly or else written badly, because there is no whole-number solution (as there must be, for this sort of problem).

So did the exercise perhaps originally state (or intend to say) that the two-digit number "is nineteen times as great as its second digit"? (Note: The "it's" is incorrect, also.) If so, then the solutions (shame, shame, gents!) provided earlier should work. Your lack of a response could indicate confusion. Where are you stuck? Please be complete. Thank you! ;)
 

lookagain

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Aug 22, 2010
Messages
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The exercise says that b is nineteen times "greater than" (+19*) that digit. So, if we are to take the exercise literally, the equation should be as follows:

. . . . .10a + b = b + 19b

... ! ;)
"(Whatever) times greater/smaller than" is wrong/poor wording from the start. It should
never be used in writing. The problem should just have been phrased as:
"Find a double-digit number which is 19 times its units digit."

Notice how I was clear that we're talking about the units digit. Someone may think
the "second digit" is by counting from the right.
 
Last edited:

Jomo

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Dec 30, 2014
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Sorry for the long absence.
Staples is correct as it does say what its says.
 
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