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3 years ago

What is the smallest number which has a reminder of 2 when it is divided by 6,7 and 8

Mathematics

1 answer:

Viktor [21]3 years ago

8 0

170 is the smallest number which has a reminder of 2 when it is divided by 6,7 and 8

<u>Step-by-step explanation:</u>

The Least or Lowest Common Multiple (LCM) or Least Common Divisor (LCD).

For two integers a and b, denoted LCM (a, b), the LCM is the smallest positive integer that is evenly divisible by both a and b. The LCM of two or more numbers is the smallest number that is evenly divisible by all numbers in the set. Let find LCM of 6, 7 and 8.

$6=2 \times 3$

$8=2 \times 2 \times 2$

$7=7 \times 1$

$L C M=2 \times 2 \times 2 \times 3 \times 7=168$

We add 2 to LCM to get the smallest number which gives a remainder of 2.

So, the number is 168+ 2 = 170

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Volgvan

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2 years ago

A recent trend in advertising is viral marketing. The goal is to convince viewers to share an amusing advertisement by e-mail or

77julia77 [94]

I'm not sure what you are looking for but I'll answer this question based on what I think is needed.

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Day 1: 130
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V(n) = 130(4)^n

Where n is the number of days since the video was first shown.

Let us assume that 10 days have past. Total view would be:

V(n) = 130(4)^n

V(10) = 130(4)^10
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2 years ago

What is the solution to this system of equations?

dalvyx [7]

There is no solution.

Step-by-step explanation:

Given:

y = 2x + 1

4x - 2y = 4

Substitute y = 2x + 1 into 4x - 2y = 4

4x - 2(2x + 1) = 4

4x - 4x - 2 = 4

0x = 4 + 2

0x = 6

Anything to the multiple of zero is zero. Hence 0x = 0, and 0 ≠ 6

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2 years ago

Can someone please help me

astra-53 [7]

Answer:

square 20 has 44 green squares

square 21 has 45 green squares

Step-by-step explanation:

To solve the problem, we need to observe the cases, and determine/define a rule for each case (odd number of sides, or even number of sides).

For square one, we note that the centre square is shared by two diagonals, so we saved one square from the two diagonals.

The side length is 3 for square 1, 4 for square 2, and so on.

Let

n= square number (1, 2,3...)

L = side length (3,4,5...)

G1(n) = function that gives the number of green squares for square n, n=odd

G2(n) = function that gives the number of green squares for square n, n=even

side length, L=n+2 ................(1)

G1(n) = twice the side length less one, as discussed above

G1(n) = 2L-1 now substitute L=n+2

G1(n) = 2(n+2) -1 simplify

G1(n) = 2n + 3

Check:

for n=1, square 1 has 2*1+3 = 5 green squares ... checks

for n=3, square 3 has 2*3+3 = 9... checks

for n=5, square 5 has 2*5+3 = 13 ....checks

For even squares, it is even easier, because

G2(n) = 2L = 2(n+2)

check:

for n=2, square 2 has 2(2+2) = 8 green squares........checks

for n=4, square 4 has 2(4+2) = 12 green squares........checks.

Fincally, we apply our formula to n=20 and n=21

square 20 : G2(20) = 2(n+2) = 2(20+2) = 44 green quars

square 21 : G1(21) = 2n+3 = 2(21)+3 = 45 green squares

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2 years ago

In a basket, 5/9 of the fruits are apples and the rest are oranges. 3/10 of the apples are green in colour. There are 15 green a

ra1l [238]

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2 years ago