Hello Teachers!

 

Largest 3 digit number divisible by 25.

 

IS there any simplest method to solve it?

 

Thanks.

you take integer multiples of 25 and see whats the largest you can get to 999.

Hello!

 

That method would take a lot of time.I want to know some short cut method.

 

I know about the method you explained to me?

 

 

Thanks.

How about 1000-25? What grade are you in hanuman_2000?

Hello Teachers!

 

Largest 3 digit number divisible by 25.

 

IS there any simplest method to solve it?

 

Thanks.

I don't know what grade level you are thinking of.

 

I suggest that you determine the largest 3 digit number. Then find the largest number less than that that is a multiple of 25, namely 25, 50, or 75.

1000 - 25.

999 is the largest

 

so divide 999 by 25 and remove the remainder (39.96 becomes just 39)

 

take that 39 and multiply it by your 25 to get your answer.

 

in this case 975 is the largest 3 digit number divisible by 25

"in this case 975 is the largest 3 digit number divisible by 25 :)"

 

Yep. It's odd how 999 can of course be divided by 25, but you wouldn't call it a number that's divisible by 25 (don't know about your English buggers and your math terms, I'm getting a severe headache even at Finnish high school grade geometry ) ).

well if it`s any consollation I HATE Maths! (I`m sure every man and his dog knows that by now), but This particular problem was just nice and comfy for my level

human, you can't say that the method is not good.

Every method to solve a simple question requires more or less the same time.

"in this case 975 is the largest 3 digit number divisible by 25 :)"

 

Yep. It's odd how 999 can of course be divided by 25' date=' but you wouldn't call it a number that's divisible by 25 (don't know about your English buggers and your math terms, I'm getting a severe headache even at Finnish high school grade geometry ) ).[/quote']

 

Divisibility is a mutliplicative property in the integers. Get used to it: 999 cannot be divided by 25 IN THE INTEGERS.

999 is the largest

so divide 999 by 25 and remove the remainder (39.96 becomes just 39)

take that 39 and multiply it by your 25 to get your answer.

in this case 975 is the largest 3 digit number divisible by 25

 

Excellent method, YT...I'm going to have to use it from now on .

well if it`s any consollation I HATE Maths! (I`m sure every man and his dog knows that by now), but This particular problem was just nice and comfy for my level

 

Interesting, though, that you gave a simple method for finding the number (which is what the poster wanted) and everyone else solved it by either brute force, or inspection (because 25 is easy) and gave a way to confirm that the answer was right.

 

If the number had been, say 37, the inspection method (which is how I did it, BTW) would likely take a bit longer.

Swansont,

 

I`m no good at maths (at all!), but when reading the question, it wasn`t like maths per se, it was just obvious, and since my perspective on maths is SIMPLE to say the least! that was the only way I could work it out (I lack the choice/will).

 

the OP presented the question nicely too, and in a way I could understand, so 10 out 10 for him too

Excellent method, YT...I'm going to have to use it from now on .

 

i thought the simple "1000 - 25" solution to be a lot simpler.

Problem is though it doesn't work in all instances.

Problem is though it doesn't work in all instances.

whe's right you know. how about changing the last 2 digits into either 00, 25, 50, 75 and +00 (the next higher hundered)

lets do it using some 'complicated' maths shall we?

 

method:

legend: X = hundreds value up.

Y = last 2 digits of course

 

let xy = 999

x=900

y= 99

so,

....................................................if y > 12

..............................................(true).......(false)

........................................if y > 37............change y to 00

...................................(true)......(false)

............................ if y > 62...........change y into 25

..........................(true)......(false)

.................... if y > 87..........then change y into 50

.................(true)......(false)

.............change y into....change y to 75

.........00 and add 1 to x

 

i made this equation so it works with real numbers, without fractions. if there is, simply round off the numbers and it should work.

 

 

substituting 999 into this metod, we get:

 

................................99 > 12

..............................(true)

........................... 99 > 37

..........................(true)

........................ 99 > 62

.......................(true)

.....................99 > 87

...................(true)

 

cahnge 99 to 00, and add 1 to y.

but wait, that would exit the limit. so we have to go i step down and change 99 into 75. so,

x = 9

y = 75

xy = 975

 

that should do it