145 1! = 1 4! = 24 5! = 120
If, as is normal, ab represents a times b, etc then ab + ab + cc = 2ab + c2 which is generally not the same as abc.
31,875,000
abc = 183.if abc + abc = cdd and d = 6, then:abc + abc + c66So, looking at the units:c + c = 6 possibly plus carry⇒ c = 3 (no carry) or c = 8 (carry)Looking at the tens:b + b + possible_carry_from_units = 6⇒ 2b + possible_carry = 62b is even, 6 is even, so the possible_carry must be even.The possible_carry is either 0 or 1; 1 is odd, so it must be 0, so c must be 3 (c = 3).This leave 2b = 6 possibly plus carry⇒ b = 3 (no carry) or b = 8 (carry)Looking at the hundreds:a + a + possible_carry_from_tens = c = 3 (as just found)⇒ 2a + possible_carry = 32b is even, 3 is odd, so the possible_carry must be odd.The possible_carry is either 0 or 1; 1 is odd, so it must be 1, so b must be 8 (b = 8).This leaves 2a + 1 = 3⇒ 2a = 2⇒ a = 1Making the number abc = 183.
Do you mean F = abc + abc + ac + bc + abc' ? *x+x = x F = abc + ac + bc + abc' *Rearranging F = abc + abc' + ab + bc *Factoring out ab F = ab(c+c') + ab + bc *x+x' = 1 F = ab + ab + bc *x+x = x F = bc
Assuming A, B, C and D all represent different digits then ABC = 183 .
ab'c + abc' + abc = a(b'c + bc' + bc) = a(b'c + b(c' + c)) = a(b'c + b) = a(c + b) I'm not sure if there's a proper name for that last step, or multiple steps to get to it, but it is intuitively correct. b + b'c is equivalent to b + c. Here's a quick truth table to show it: bcb'b'cb'c+bb+c0010000111111000111100 11
i think its ABC and if that not right ask a teacher for the answer
145 1! = 1 4! = 24 5! = 120
You want: abc + ab Factor out the common terms which are "a" and "b" ab ( c + 1 )
If, as is normal, ab represents a times b, etc then ab + ab + cc = 2ab + c2 which is generally not the same as abc.
'abc', 'acb', and 'cba' are all the same number. 459+495=954 so A is 4 B is 5 and C is 9
31,875,000
a3 + b3 + c3 + 2(a2)b + 2(b2)c + 2(a2)c + 2ab2 + 2(c2)b +abc
In this case you want to group the terms so they have at least two terms in common. First step group and rewrite it: abc + a'bc + a'b'c' + a'b'c + ab'c' + abc' = Use the rule Identities x(y+z)=xy+xz: bc(a+a') + a'b'(c'+c) + ac'(b'+b) = Use the rule Identities x+x'=1: bc (1) + a'b'(1) + ac'(1) = Use the rule Identities x(1) = x: bc+a'b'+ac'
A x B x C = ABC
abc = 183.if abc + abc = cdd and d = 6, then:abc + abc + c66So, looking at the units:c + c = 6 possibly plus carry⇒ c = 3 (no carry) or c = 8 (carry)Looking at the tens:b + b + possible_carry_from_units = 6⇒ 2b + possible_carry = 62b is even, 6 is even, so the possible_carry must be even.The possible_carry is either 0 or 1; 1 is odd, so it must be 0, so c must be 3 (c = 3).This leave 2b = 6 possibly plus carry⇒ b = 3 (no carry) or b = 8 (carry)Looking at the hundreds:a + a + possible_carry_from_tens = c = 3 (as just found)⇒ 2a + possible_carry = 32b is even, 3 is odd, so the possible_carry must be odd.The possible_carry is either 0 or 1; 1 is odd, so it must be 1, so b must be 8 (b = 8).This leaves 2a + 1 = 3⇒ 2a = 2⇒ a = 1Making the number abc = 183.