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# 54 is 144% of what number?.

54 is 144% of what number?

## Research, Knowledge and Information :

### What is 144 percent of 54 - step by step solution

What is 144 percent of 54 ... We assume, that the number 54 is 100% ... 54/x=100/144 (54/x)*x=(100/144)*x ...

### 144 (number) - Wikipedia

144 is the smallest number with exactly 15 divisors, but it is not highly composite since a smaller number 120 has 16 divisors. 144 is divisible by the value of its ...

### 54 is 150 percent of what number - Answers.com

54 is 150 percent of what number? SAVE CANCEL. already exists. Would you ... conversions and physics are my interests on Answers.com. What number is 35 percent of 54?

### What is the cube root of 54? | Reference.com

The cube root of a number, x, is equal to the number, y, such that y times y times y equals x. For the number 54, ... The cube root of 54 is equal to 3.78.

### 144% of what number is 126? - OpenStudy

144% of what number is 126?what do you think?. 126/144?. 144% * n = 126 144/100 * n = 126 1.44 * n = 126 n = 126/1.44. exactly^. oh okay th ... ...

### 5.4 Solving Percent Problems Using the Percent Equation

5.4 Solving Percent Problems Using the Percent Equation ... Suppose we know that 68% of a number is 204. The percent is 68% = 0.68, converted to a decimal.

### Percent word problem: 78 is 15% of what number? (video ...

In this example, you working with us to find the number that is expressed as a given percentage. ... Percent word problem: 78 is 15% of what number?

### List of numbers - Wikipedia, the free encyclopedia

A list of articles about numbers (not about numerals). Topics include powers of ten, notable integers, prime and cardinal numbers, and the myriad system.

## Suggested Questions And Answer :

### use the integers 0, -2, -3, -4, -5, -6, -7, -8, and -10 to fill in a 3x3 magic square

First, find out what the sum is. Consider 3 rows of 3 numbers: each row has sum S. The rows contain all the numbers, so since the sum of the numbers is -45, 3S=-45 and S=-15. Take each number in turn and work out what other numbers can go with it to make -15: 0: (-10,-5),(-8,-7) -2: (-10,-3),(-8,-5),(-7,-6) -3: (-10,-2),(-8,-4),(-7,-5) -4: (-9,-2),(-8,-3),(-6,-5) -5: (-10,0),(-8,-2),(-7,-3),(-6,-4) -6: (-7,-2),(-5,-4) -7: (-8,0),(-6,-2),(-5,-3) -8: (-7,0),(-5,-2),(-4,-3) -10: (-5,0),(-3,-2) Now, consider the square: C1 C2 C3 C4 C5 C6 C7 C8 C9 Each cell participates in various sums. C1, for example, is in the first row and column and on a diagonal, so it is involved in 3 sums. C3, C7 and C9 are also similarly involved. Here's a complete list: C2,C4,C6,C8: 2 sums C1,C3,C7,C9: 3 sums C5: 4 sums Now we match the number of sums with the numbers that can be involved in at least the same number of sums. C5=-5. 2 sums: {C2 C4 C6 C8}={0 -6 -10} (exact) < {-2 -3 -4 -7 -8} (3 sums) 3 sums: {C1 C3 C7 C9}={-2 -3 -4 -7 -8}, one of these can be used to satisfy 2 sums, leaving 4 to match the cells. So {C1 C3 C7 C9}={-2 -3 -4 -7)|{-2 -3 -4 -8}|{-2 -3 -7 -8}|-2 -4 -7 -8}|{-3 -4 -7 -8} and {C2 C4 C6 C8}={0 -6 -8 -10}|{0 -6 -7 -10}|{0 -4 -6 -10}|{0 -3 -6 -10}|{0 -2 -6 -10} Because of symmetry we can assign 0 to C2. So C8=-10: C1 0 C3 C4 -5 C6 C7 -10 C9 C7+C9=-5, which means only -2 and -3. Because of symmetry we can assign either number to C7: C1 0 C3 C4 -5 C6 -2 -10 -3 C1-8=-15 so C1=-7; C3-7=-15, so C3=-8: -7 0 -8 C4 -5 C6 -2 -10 -3 C4=-6 and C6=-4: -7 0 -8 -6 -5 -4 -2 -10 -3

### Genet multiplied a 3-digit number by 1002 and got AB007C, where A, B, and C stand for digits. What was Genet's original 3-digit number?

Write the three digit number as 100a+10b+c, where a, b and c are the digits. Write 1002 as 1000+2, now expand (100a+10b+c)(1000+2)= 100000a+10000b+1000c+200a+20b+2c=100000A+10000B+70+C. I found it easier to equate digits by comparison by arranging this sum as a layout in long multiplication: ............................................1 0 0 2 ...............................................a b c ...........................................c 0 0 2c ........................................b 0 0 2b 0 .....................................a 0 0 2a 0 0 .....................................A B 0 0 7 C We should be able to equate the digits by position by comparisons. C must be an even number 0, 2, 4, 6 or 8. But in the tens we have 7, so c is 5, 6, 7, 8 or 9, to produce a carryover converting 6 to 7 in the tens. That tells us that b must be 3 or 8 because 2*3+1=7 or 2*8+1=17. But in the hundreds we have 0, so b can't be 8 because there would be a carryover, so b=3. And we can see that 2a must be a number ending in zero, so a=0 or 5. But a can't be zero because we would have a 3-digit number starting with zero making it a 2-digit number so a=5, and 2a=10, which is a carryover to the thousands where c is. But the thousands digit is zero, so c=9. We have all the digits: a=5, b=3 and c=9 making the number 539. 539*1002=540078.

### seven times a two digit number

Short answer:  The two digit number is 36. Long answer: 7 * x1x2 = 4 * x2x1 "If the difference between the number is 3. . ." Last digits: x2: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 7*x2:  0, 7, 4, 1, 8, 5, 2, 9, 6, 1 x1:  0, 1, 2, 3, 4, 5, 6, 7, 8, 9 4*x1:  0, 4, 8, 2, 6, 0, 4, 8, 2, 6 The only way this works is when these last digits for 7*x2 and 4*x1 are the same.  That means the 7*x2 line can only be: 7*x2:  0, 4, 8, 2, 6 Which means the possible values for x2 are: x2:  0, 2, 4, 6, 8 Now let's look at x1.  Right now the possible values for x1 are: x1:  0, 1, 2, 3, 4, 5, 6, 7, 8, 9 But we have to end up with a choice from x1 and x2 having a difference of 3 (odd number).  There is no way to get an odd number by subtracting an even number from an even number.  That means x1 has to be odd.  Our possible values for x1 are now: x1:  1, 3, 5, 7, 9 And our possible values for x2 are: x2:  0, 2, 4, 6, 8 The possible combinations for x1x2 and x2x1 are: 10, 01 12, 21 14, 41 16, 61 18, 81 30, 03 32, 23 34, 43 36, 63 38, 83 50, 05 52, 25 54, 45 56, 65 58, 85 70, 07 72, 27 74, 47 76, 67 78, 87 90, 09 92, 29 94, 49 96, 69 98, 89 We want 7*x1x2 = 4*x2x1, so we can do 7 * the first column and 4 * the second column: 70, 4 84, 84 98, 164 112, 244 126, 324 210, 12 224, 92 238, 172 252, 252 266, 332 350, 20 364, 100 378, 180 392, 260 406, 340 490, 28 504, 108 518, 188 532, 268 546, 348 630, 36 644, 116 658, 196 672, 276 686, 356 But since we want 7*x1x2 to equal 4*x2x1, that list reduces to: 84, 84 252, 252 The corresponding x1 and x2 values are: 12, 21 36, 63 But the difference between x1 and x2 is 3, so we can't use x1 = 1, x2=2.  We have to use x1 = 3, x2 = 6. Answer:  The two digit number is 36. Check:  7 * 36 = 4 * 63 252 = 252 good. 6 - 3 = 3 good.

### How many different distributions can the manager make if every employee receives at least one voucher?

There must be 100 vouchers because each is worth RM5 and the total value is 500 ringgits or RM500. (i) Each employee receives at least 1 voucher, so that means there are 95 vouchers left to distribute. We can write each distribution as {A,B,C,D,E}: starting with {0,0,0,0,95}, then {0,0,0,1,94}, {0,0,0,2,93}, ..., {0,0,0,95,0}, {0,0,1,0,94}, ..., {0,0,95,0,0}, ..., ..., {95,0,0,0,0}. So when A=B=C=0, {D,E} range from {0,95}, {1,94}, ..., to {95,0}, 96 ways. When A=B=0 and C=1, {D,E} range from {0,94} to {94,0}, 95 ways. Finally when C=95 so {C,D,E}={95,0,0} we will have covered 96+95+...+1=96*97/2=4656 ways. (The sum of the whole numbers from 1 to n is given by S=n(n+1)/2.) That was for B=0; when B=1, {C,D,E} ranges from {0,0,94} to {94,0,0} to cover 95*96/2=4560 ways. When B=2 it's 94*95/2=4465 ways. So for A=0 we have 4656+4560+4465+...+3+1 = 96^2+94^2+...+2^2 = 4(48^2+47^2+46^2+...+2^2+1^2)=4*48*49*97/6=152096 (the sum of the squares of the whole numbers from 1 to n is given by S=n(n+1)(2n+1)/6. Also, the sum of whole numbers between 1 and n taken in pairs gives us: n(n+1)/2+(n-1)n/2 for each pair. This is n^2/2+n/2+n^2/2-n/2=n^2. For the next pair we get (n-2)^2 and so on.) That was just for A=0! For A=1 we have {1,0,0,0,94} to {1,94,0,0,0}. This will give us 4560+4465+...+10+6+3+1 = 95^2+93^2+...+5^2+3^2+1. There is a formula for this sum. It is S=(n+1)(2n+1)(2n+3)/3, where 2n+1=95, so n=47. So for A=1, the number of ways is 48*95*97/3=147440. For A=2 we have {2,0,0,0,93} to {2,93,0,0,0} which produces 94^2+92^2+...+4^2+2^2 = 4(47^2+46^2+...+1) = 4*47*48*95/6 = 142880. So we alternate between two formulae as A continues to go from 3 to 95.  More to follow...

### How many numbers can be made with 3 beads and 3 columns, and then 4 beads and four columns

Consider the different combinations of 3 beads: 3 together, 3 all separate, and a bead and a pair of beads. Now the columns. With 3 columns we can choose which column to put the 3 beads together in; that gives us 3 different numbers (003, 030, 300). Separate all the beads, one per column, that's another number (111). That leaves us with a single bead and a pair, i.e., the digits 0, 1 and 2. There are 6 ways of permuting three different objects. Add this to the other 4 and we get 10, so there are 10 numbers possible using three beads and three columns. With 4 beads, all kept together and 4 columns, there are 4 numbers. Separate the beads and put one in each column and we get one more number. That leaves us with two pairs, and a single bead and 3 together. The numbers made of two pairs consist of the digits 0, 0, 2 and 2. If this had been four different digits we would have 24 different permutations, but we only have two different digits, so we need to reduce the permutations by a factor of 2 twice, so 24/4=6. Finally we have the digits 0, 0, 1 and 3. If this had been just 0, 1 and 3, i.e., three different digits it would be similar to the 3 bead problem and there would be 6 ways of permuting these digits. The extra 0 can be at the beginning or end of each of these 6 permutations making 12 altogether. Another way of looking at it is to reduce the 24 permutations of 4 different objects by a factor of 2 to compensate for two digits (0) being the same. So to summarise by adding together all these in order we have 4 (all beads together) + 1 (all beads separate) + 6 (two pairs) + 12 (single bead and 3 beads together) = 23 different numbers.

### What are the next three numbers in the sequence 10, -1, 0

There aren't enough starting numbers to figure out the pattern. Or, more correctly, there are many patterns that could fit those starting numbers.  The problem is, with this few starting numbers, we can't identify the one specific pattern that gives us these starting numbers. Example: 10, -1, 0 Pattern: subtract 11, add 1, subtract 11, add 1 Result:  10, -1, 0, -11, -10, -21, -20, . . . . Example: 10, -1, 0 Pattern: divide by -10, add 1, divie by -10, add 1 Result:  10, -1, 0, 0, 1, -1/10, 9/10, -9/100, 91/100, . . . . Example: 10, -1, 0 Pattern: square the number and subtract 101, then add 1, then square the number and subtract 101, then add 1 Result:  10, -1, 0, -101, -100, -10101, -10100, . . . . All of these examples are patterns that generate 10, -1, 0, so all ofthese examples are accurate given 10, -1, 0 as a starting pattern

### how many ways are there to add and get the sum of 180

There are an infinite number of ways to get 180 from two numbers, if we count decimals and fractions as well as other real numbers; but if we are limited to positive integers greater than zero and just the sum of two of them, we are limited to x and 180-x. If we also exclude 90+90 because the numbers are the same, then we have 1 to 89 combined with 179 to 91, which is 89 pairs. Moving on to the sum of three different numbers, let's make 1 plus another two different numbers adding up to 179. So we have 2+177, 3+176, ..., 87+92, 88+91, 89+90, which is 88 groups combined with 1. Move on to 2 plus another two different numbers adding up to 178: 3+175, ..., 87+91, 88+90, which is 86 groups. Then we move on to 3 plus 177: 4+173, ..., 86+91, 87+90, 88+89, 85 groups. And so on, with reducing numbers, until we get to 59, 60 and 61. Let's divide the numbers into two groups A and B. In A we start with 1 and in B we put 2 and (180-A-B)=177 as a pair (2,177). Then we put the next pair in group B: (3,176), then (4,175) and keep going till we have used up all the numbers, ending up with (88,90). Then we count how many pairs there are in group B and pair it up with the number in group A, so we start with (1,88) which covers all the combinations of numbers in group B. Now we move to 2 in group A, put all the pairs adding up to 178 in group B, and finally put the count of these pairs with 2 in group A: (2,86). We then move on to 3, and so on, putting in the counts to make up the number pair in group A. When we've finished by putting the last count in group A, which is (59,1), we can forget about group B and look at the pattern in group A. What we see is this: (1,88), (2,86), (3,85), (4,83), (5,82), (6,80), (7,79), ... See how the counts come in pairs with a gap? All the multiples of 3 are missing in the counts sequence (e.g., 87, 84, 81). We find there are 29 pairs and one odd count, 88, which is unpaired. Number the pairs 0 to 28 and refer to the pair number as N. Add the counts in the pairs together so we start with pair 0 as 86+85=171, pair 1 as 165, pair 2 as 159, and so on. The sequence 171, 165, 159, ..., 3 is an arithmetic sequence with a start of 171 and a difference of 6 between each term in the sequence. [Note also that the terms in the series are all multiples of 3: 3*57, 3*55, 3*55, ...] The rule for the Nth term is 171-6N. When N=0 we have the first term 171 and when N=28 the last term is 3. There is one more term at the end which is unpaired made up of the numbers 59, 60 and 61. We can combine this with the unpaired (1,88). We can find the sum of the terms in the series, which will tell us how many ways there are of adding three different integers so that their sum is 180 (like the sum of the angles of a triangle).  To find the sum of the terms of the series we note that there are 29 terms (0 to 28) and they all contain 171, so that's 171*29=4959. We also have to subtract 6(0+1+2+3+...+28)=6*28*29/2=2436. So 4959-2436=2523. [The sum of the series is also 3(57+55+53+...+5+3+1)=2523.] To this we add the "odd couple" 88+1=89 and 2523+89=2612. Add also the 89 which is the number of pairs of integers adding up to 180 we calculated at the beginning. The total so far is 2612+89=2701 ways of adding 2 or 3 positive integers so that their sum is 180. If you want to go further, please feel free to do so!

### Math Word Problem with Graph

Plot two lines on the same graph: A=50-C (constraint on numbers of hair bands) and A=(75-3C)/5 (from 5A+3C=75, the flower constraint), where A=number of adult bands and C=the number of child bands. The area under the first graph (a straight line joining the A intercept 50 and C intercept 50) includes the area under the second graph (A intercept 15, C intercept 25), so the more severe constraint is the second graph. Its area is enclosed by the axes and the line. The income equation R (revenue)=5A+3C matches the flower constraint, as it happens. No matter how many bands of either type are made the income will always be \$75 if all the available flowers are used. So 15 adult bands will fetch 15*5=\$75 as will 25 child bands=25*3=\$75. If we take 9 adult bands, for example, we need 45 flowers, leaving 30 flowers from which 10 child bands can be made. The income is \$45+\$30=\$75. The total number of bands is 19, well below the capacity of 50. The number of child bands will always be a multiple of 5 and the number of adult bands a multiple of 3 to maximise income at \$75 (C=0, 5, 10, 15, 20, 25 with A=15, 12, 9, 6, 3, 0, so A+C=15, 17, 19, 21, 23, 25). The graph vertices are defined by the line A=(75-3C)/5 as (C,A)=(0,15), (0,0) and (25,0). These vertices enclose an area which is enclosed by the vertices (0,50), (0,0) and (50,0). These cover the constraints outlined in the question.