The hcf and lcm of two numbers are 52 and 624 respectively. One number is 156 find the other
Answers
Answer:
THE OTHER NUMBER IS 12
Step-by-step explanation:
Answer:
The sum of two numbers is 52 and their LCM is 168. What are the numbers?
H.c.f = h.c.f of (52,168)
= 4
Lcm / hcf
= 168/4
= 42
Write in co prime
14*3 or 6*7
So h.c.f *(one coprime part)
4*14 & 4*3
56 & 12
Or
4*6 & 4*7
24 & 28
But here its given that sum should be 52
So 24 & 28 is the answer
The numbers are 24 and 28. But how did I get it?
Simple.
Let us assume that the two numbers are a and b. Since LCM is an even number, this means atleast one of a and b has to be an even number. However, if one of them is even, the other should be even too because 52(even) minus even is even.
Therefore both a and b are even.
Wow we can take two numbers between 2 and 50 which are even, their sum is 52 and their multiples is 168.
You will get 24 and 28.
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52=4*13; 168=4*6*7 So HCF=4
Let the numbers be 4a and 4b
4a+4b=52; a+b=13
4ab=168; ab=42
a-b=square root of {(a+b)^2–4ab}=of{169–168}=of 1
=plus or minus 1
So a=6 and b=7;
The numbers are 24 and 28
The sum of two numbers is 320 and their LCM is 1584. What are the numbers?
The sum of two numbers is 144 and their LCM is 420. What are the numbers?
The L.C.M. of two numbers is 12 and their sum is 10. What are the numbers?
24 and 28. Here's how I got to it...
First factorize 168...
168 = 2 * 2 * 2 * 3 * 7
Now from the set of numbers { 2, 2, 2, 3, 7 }, choose two subsets such that the sum of their products is 52.
By trial and error, you will get two subsets {2, 2, 2, 3} and {2, 2, 7} such that
24 + 28 =52 !
As LCM of two numbers is 168 that two numbers must be factors of 168
now think about what are all factors of 168 that are less than 52(because sum of those two is 52)
factors=(2,3,4,6,8,7,8,12,14,24,28,42)
from above factors u have to select two numbers such that sum of two is 52
only possibility is 24+28
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168=2×2×2×3×7 ,Let number are a , 52-a ie a<52
168 is divisible by a and 52-a
if 168=a×2 , then a=84 , not possible
if 168=a×3 , then a=56 , not possible
if 168= a×7 , then a=24 , only possible value of a
So the numbers are 24 and 52–24=28
numbers 24,28
Let the numbers be x and y.
x + y = 52
The prime factors of 168 are:
2 x 2 x2 x3 x7
With this information by hit or miss we can find out the two numbers as 24 and 28 whose total is 52.