Challenge Problem: Each Letter stands for a digit. Find out what number each letter stands for using the equation below. C + C + C + C + R = FF F * R= MF C * C = MR So C= ? R= ? F= ? M= ? (fourth grade challenge; trying to help my little sister)
anybody? @timo86m @shamil98 , @Zale101 @iambatman
well, is there a number at least given ? that fourth grade work seems challenging lol
Nope no numbers lol
Are you sure no numbers were given o_O
nope no numbers, want a picture for proof ? lol
yesh :3
you idiot you don't believe me. lmao gimme a sec.
lol
I believe you
This is one of those problems where the answer is right in front of you but you end up over complicating it.
yupp, what if they all equal 1 ?I mean if they do, it'll make sense
I feel sick
And if Fiz gets this answer i'm going to be pissed
Hmm have you tried assigning C, R F,and M a value? C + C + C + C = C^4 C^4 + R = F if F were 18, c was 2 and r was 2 then.. 2^4 + 2 = 18 but they would also have to match the other problems. Try finding a number that satisfy all of the problems >.<
;-; if fiz got it right I would also be shocked >.<
Yeah I'm thinking you have to assign the digits yourself XD
o.0
LOl piss him off.
Waiiit c + c + c + c isn't c^4 >_<
that would onyl be c * c * c * c >_<
only*
Sheit I messed up on adding them ;-; im terra bad U_U you get the idea though
The proof. @fiz you're making sense but that is too much to solve for a 4th grader. I think it is a simple equation just very confusing.
an acceptable answer is probably C = 1, R = 1, M = 1 and F = sqrt(5)
oh dang square roots, she haven't learned that, but I see what you're doing
Pfft, this is just adding >_< adding roots *cough* seems complicated for a 4th grade -looks away- lol but seriously just find some values... I think that FF is a double digit meaning like FF like 22 Or f*f which if it resulting in 22 divide by 2 and get 11. So each f = 11 And then you plug in 11 for the other f's and find a R value for example :D
@mathmale I'm not making sense honestly... I'm multi-tasking D:
just reposting correct version second row: R = M third row: C^2 = R^2 = M^2 C = R = M first row: 5C = F^2 equally, 5R = F^2 and 5M = F^2 \[F = \sqrt{5C} = \sqrt{5M} = \sqrt{5R}\] where C, R and M are any number except 0 (because we divided by F to make our assumption in the second row; therefore F can't equal zero
do fourth graders learn about square roots?
Whoopwhoop got to give away two medals this time. See, if there was no M this would be much easier. D; But I'll take what you both did and figure it out with her.
nope I asked her, she has no idea what they are
k
This is not 4th grade math o_O
Maybe for an alien race
and batman seen the proof lol ?
YES it is. it's a challenge homework
i think her teacher assigned the class the numbers or something
Thanks Fizi (:
oh wow. EACH LETTER STANDS FOR A DIGIT ignore my answer
*Cough* yw
LOL if her teacher gave her the numbers and she comes with a long equation high schooled based lmao her reaction X_X
@ zale ; nope, lol euler @tHe_FiZiCx99 haha quit coughing.
Prob why she's teaching grade 4 math
err he/she
lol! yea this is impossible. especially or fourth grade
for*
I repeat: This is one of those problems where the answer is right in front of you but you end up over complicating it.
I'll just put all equal 1, and tomorrow she'll ask the teacher, ill post what was the answer. lol
baman I would have seen it >_< like im just op :*
I'm more OP, dem gadgets
Guys 1 won't work Dx
WTH.
1 is too easy >_<
and this question is too confusing. I thought maybe it could be one since this is 4th grade.
I know how to do it
How??
Brb
alright
ill bbs and try it later
Lol tagg me if you ever got it.
Is it a word problem? C+C+C+C+R = Foreseer
C=4, R = 6, F = 2, M = 1
Nope it just says FF
OMG you're are brilliant! At first I thought ( ff means f * f) I plugged it like this 4 + 4 + 4 + 4 + 6 = 2 x 2 0r 4 2 x 6 = 1x2 4 x 4 = 1x 6 But now I see 4 + 4 + 4 + 4 + 6 = 2 2 2 x 6 = 12 4 x 4 = 1 6 Haha Thank you so much (:
whpalmer, how did you solve this?
Oh Timaa lol, but yup @whpalmer4 is correct :P
Took fiz's medal away and gave it to whp
i suppose is trial and error then
not quite, I'll write up my method, just a minute...
So, starting with the C + C + C + C + R = F F or 4C + R = F*10 + F trying out the various values of C from 0 to 9, and then adding the possible values of R from 0 to 9 to that, and looking for a sum that has two identical digits, the possibilities are: C = 8 R = 1 F = 3 C = 5 R = 2 F = 2 C = 2 R = 3 F = 1 C = 7 R = 5 F = 3 C = 4 R = 6 F = 2 C = 1 R = 7 F = 1 C = 9 R = 8 F = 4 C = 6 R = 9 F = 3 Now let's discard all the cases where we have the same digit assigned to multiple letters: C = 8 R = 1 F = 3 C = 2 R = 3 F = 1 C = 7 R = 5 F = 3 C = 4 R = 6 F = 2 C = 9 R = 8 F = 4 C = 6 R = 9 F = 3 Our next constraint is C*C = M*10+R Going through our candidates: 8*8 = 64, but R = 1, not 4 2*2 =4, but R = 3, and we don't have a tens digit, 7*7 = 49, but R = 5, not 9 4*4 = 16, R = 6, good, and implies M = 1 9*9 = 81, R = 8, not 1 6*6 = 36, R = 9, not 6 So C = 4, R = 6, M = 1, and F = 2
I still find it surprising they would assign this to a 9 year old lol.
at first, i thought i was dealing with roman numerals haha...guess not... awesome solution dude
Yeah, I haven't met anyone at my kid's school who could have done this at age 10 or 11, I think, much less at 9.
you dont need to test all r's only 1,4,6,9,0
solution is unique.
and r equal 4 also implies in M=6
See, I kind of did what you did except I kept thinking FF = F*F, MR = M*R, & MF = M*F So I never really got the solution. Thanks for explaining it really well, I showed my sister this morning she got all excited when it was solved. lol & Batman me neither, but the teacher put it for her as a challenge/ bonus grade for fun I guess. It wasn't something they had to do.
I thought it was either f*f or a double meaning like 2 number >_<
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