Find the smallest number which when divided 18,12,24 leaves a remainder of 4 in each case
Answers
Step-by-step explanation:
Find the smallest number which when divided by 18,12,24 leaves a remainder of 16,10 and 22 respectively. 72 is completely divisible by 18,12, 24. so, minus the remainder from the numbers and the answer is two.
hope you understand..
Answer:
Step-by-step explanation:
The key thing with maths is that you need to understand the concepts first a bit more deeply before embarking on their calculations - solutions. So lets first start there.
Are you aware of this notation?
d N+r={ d⋅n+r ∖ n∈N }
so
d N+r={ d⋅0+r , d⋅1+r , d⋅2+r ,… }
Now, the nice part is that all natural numbers that, divided by the positive integer d ≠0 leave a remainder r , with d >r∈N , belong to the set d N+r . Actually, this is their set!
To rephrase: the set d N+r is the set of all (non-negative) integers that when divided by (the positive integer) d (the divisor) produce (non-negative integer) r (the remainder).
With this notation, rethink your question. The numbers you seek form sets of the form above. For example, the numbers which, divided by 18, give 16 as the remainder, form the set
18 N+16={18⋅n+16∖n∈N}
={0⋅18+16,1⋅18+16,2⋅18+16,…}
={16,34,52,70,88,106,124,142,160,178,196,214,232,…}
Likewise, the numbers which, divided by 12, give 10 as the remainder, form the set
12 N+10={12⋅n+10∖n∈N}
={0⋅12+10,1⋅12+10,2⋅12+10,…}
={10,22,34,46,58,70,82,94,106,118,130,142,154,166,178,190,202,214,226,…}
and, last, the numbers which, divided by 24, give 22 as the remainder, form the set
24 N+22={24⋅n+22∖n∈N}
={0⋅24+22,1⋅24+22,2⋅24+22,…}
={22,46,70,94,118,142,166,190,214,238,…}
24 N+22
So, to solve your problem, first you need to find the intersection of the above sets, and then seek its least element.
Their intersection set can be easily found if you find the common elements of all three sets. In our case, these are 70, 142, 214, etc.
To answer your initial question, the first element of the intersection we found above is 70.
Now, can we see how all numbers that satisfy your above requirements look like?
Well, they look like this:
70,
142=72+70=72⋅1+70,
214=142+72=(72+70)+72=72⋅2+70,
286=214+72=(72⋅2+70)+72=72⋅3+70, etc.
so the numbers are of the form 72⋅n+70 with n∈N . But this is the set
72 N+70={72⋅n+70 ∖n∈N }
={0⋅72+70,1⋅72+70,2⋅72+70,…}
={70,142,214,286,…}
This is the set of all numbers that satisfy your demands, and its least element is what you want. But if you need any other such numbers, they are all there, neat and sweat.
If you need negative numbers that satisfy the same demands, simply don’t restrict the n variable to the positive integers, but let it roam freely in all integers. Then your set would be
72 Z+70={72⋅k+70 ∖k∈Z }
={…,−2⋅72+70,−1⋅72+70,0⋅72+70,1⋅72+70,2⋅72+70,…}
={…,−290,−218,−146,−74,−2,70,142,214,286,…}
All the math with congruences, modular arithmetic etc. that you learn, deep down have to do with sets like these above and how they relate to each other. These sets are equivalence classes modulo a number.
I believe that knowing that you play with sets instead of trying to calculate a number, makes the whole thing more fun. And it relates heavily to abstract algebra later on…
3.3K viewsView 1 Upvoter
please brainlist and like