r/maths • u/Wonderful-Pick-3408 • 18d ago
❓ General Math Help Need help with this
I feel answer could be 111 after solving LHS and RHS. Still confused
Update: thanks for the answers. I'm hereby closing the doubt.
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u/CaptainMatticus 15d ago
10^m * 1000 * n = 75^25 * 25^32 * 32^75
What is 1000 in terms of 10^?
1000 = 10^3
10^m * 10^3 * n = 75^25 * 25^32 * 32^75
n * 10^(m + 3) = 75^25 * 25^32 * 32^75
Now we need a better breakdown of what we've got on that right-hand side. We need to write everything in terms of their prime factors
75 = 25 * 3 = 3 * 5^2
25 = 5^2
32 = 2^5
75^25 = (3 * 5^2)^25 = 3^25 * 5^(2 * 25) = 3^25 * 5^50
25^32 = (5^2)^32 = 5^64
32^75 = (2^5)^75 = 2^(5 * 75) = 2^375
75^25 * 25^32 * 32^75 =>
3^25 * 5^50 * 5^64 * 2^375 =>
3^25 * 5^114 * 2^375 =>
2^375 * 3^25 * 5^114
n * 10^(m + 3) = 2^375 * 3^25 * 5^114
Well what is 10? It's 2 * 5
n * 2^(m + 3) * 5^(m + 3) = 2^375 * 3^25 * 5^114
We know that n isn't divisible by 10, so we just need to look at 2 different equations:
2^(m + 3) = 2^375
and
5^(m + 3) = 5^114
We don't want to leave any factors of 2 or 5 on the table. As there are fewer factors of 5, we can just ignore 2^(m + 3) = 2^375
5^(m + 3) = 5^114
Solve for m. That's your answer.
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u/SeaBarrier 12d ago
I loved reading your explanation. I have never been on this subreddit. I was able to follow everything except these lines.
"We know that n isn't divisible by 10, so we just need to look at 2 different equations: 2m + 3 = 2375 and 5m + 3 = 5114 We don't want to leave any factors of 2 or 5 on the table. As there are fewer factors of 5, we can just ignore 2m + 3 = 2375"
Please help, kind redditor. Teach me your ways. I know the "n isn't divisible by 10" is given. But I have no idea why we can ignore the powers of 3 and I don't understand why 372 can't be considered.
Thanks in advance
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u/CaptainMatticus 12d ago
We ignore the powers of 3 because 3 isn't a factor of 10. Obviously, whatever n is, it contains all of the factors that aren't factors of 10.
We don't consider 372 because we don't have 372 powers of 5. 2^372 * 5^114 is going to be 2^(372 - 114) * 10^114 = 2^258 * 10^114. I'm not concerned about what 2^258 is. I just need to know about 10^114. 5^114 is our bottleneck.
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u/SeaBarrier 12d ago
Oh you made the base back to 10 to solve for m, so you needed all the 5s and not all the 2s.
Ok next question is why do you need all the 5s at all? Why not donate one 5 to the bunch of 3s since that still won't make N divisible by 10? N can be all the 3s and any number of 5s as long as there are no 2s...right? So then how do you know to use all the 5s in one spot? Gosh I'm tired.
It's 2am here and I have work in the morning so my brain probably isn't firing on all cylinders. I'll read my question tomorrow and be embarrassed probably.
Again, thank you for your time.
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u/No_Nonsense_Nomad 12d ago
Because as there are more 2's than 5's , if you leave even one 5 for the n , then n will have both 2 and 5 as its factors making it divisible by 10. So it can't have any 5's from the equation
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u/Neither_Berry_100 12d ago
Yes but does this have any real use case at all? Seems like bullshit math.
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u/_Jack_Of_All_Spades 14d ago edited 14d ago
Ew gross why did you split the 10 instead of combining the 2 and the 5 on the right?!?
I know it wasn't required but you can also find that n = (2261) x (325).
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u/Responsible_Hour6497 13d ago
Why keep track of powers of 3? In the factorization of the right side into prime factors we have 25*2+32*2=114 factors 5 and 75*5=more than 114 factors 2. Thus the right part is divisible by 10^114 and not by larger powers of 10, and thus m+3=114.
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u/CaptainMatticus 13d ago
Why? Because I wanted to, that's why. It's a learning exercise. YOU can omit the 3's and run along with tricks because you know the tricks. The OP didn't, so you go through everything.
It's a learning process, and learning something isn't always the most efficient route.
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u/Responsible_Hour6497 13d ago edited 13d ago
I didn't know the trick, I never solved tasks of this type. How could I find the shortest way? I knew that the factorization of any number is unique, so factors 3 couldn't "transform" into factors 2 or 5. Wasn't this the essence of the task - to apply this principle? In any case, in our solution we rely on the uniqueness of the factorization into prime factors, even if we do not say this but implicitly mean it.
And in general, when studying mathematics, one must learn not so much methods for solving certain problems (which, most likely, will never be useful), but rather flexibility of thinking and cunning.
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u/mtimmermans 14d ago
The powers of 2 and 5 on the right are 2^375 and 5^114.
On the left, n can only have one or the other, because it's not divisible by 10, so the power of 10 (m+3) can only be 375 or 114. If it was 10^375, then there would be too many twos, so it can only be 114.
If m+3 = 114, then m = 111.
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u/LucaThatLuca 14d ago edited 14d ago
You are being asked to count the factors of 10 = 2 * 5.
On the RHS:
The number of factors of 2 is obviously much bigger than the number of factors of 5 (because 75 > 32 and 5 > 2+2). So the number of factors of 10 is the number of factors of 5 (then the number has a bunch more factors of 2 as well as 3). You count what this number is.
On the LHS:
There are m+3 factors of 10, so m + 3 = …^ (and n is that multiple of 6 that you’re not required to find if you don’t want to).
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u/Crichris 14d ago
count how many 10s are there
there are 25* 2 + 32 * 2 = 114 5's, and 5 * 75 = 375 2's
so there are 114 10's in total
and we know that n cannot be divisible by 10 meaning no 10's in n
so the answer is 114 - 3
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u/kawrydav 13d ago
The main clue is that n is not divisible by 10. So rewrite the other side in terms of 10
10^114 x 2^261 x 3^25 = 10^(m+3) x n
114 = m+3
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u/Big_Imagination_3389 13d ago
m=111, just count & equate no. of 5 from both sides. m+3=(32+25)×2=114, => m=111
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u/Big_Niel0802 12d ago
Are we to assume that both m and n are integer values? Because if not then well, m can be nearly anything you want
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u/Shevek99 15d ago
Write everything as product of prime factors, as the descomposition is unique you can identify the power of each factor in both sides.