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I want to design a circuit that multiplies a number by 6 an input binary number using only 4bit full adders. I designed the combinational logic but couldn't do it with 4bit adders as i were asked in my exam. any help?

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  • \$\begingroup\$ What is the input width? You might want to start by figuring out which bits in the output depend on which bits in the input. \$\endgroup\$ – Chris Stratton May 28 '16 at 1:56
  • \$\begingroup\$ the input is only one digit (4 bits) \$\endgroup\$ – iMohaned May 28 '16 at 1:57
  • \$\begingroup\$ I have considered the 6(0110) as the multiplier and the input(wxyz) as the multiplicand and made the mathematical operation with my hand and I have found that we need to add two numbers (wxy+wxyz) if i implement that with BCD adder , will I get the right answer? \$\endgroup\$ – iMohaned May 28 '16 at 2:04
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Given you only need to multiply by 6, rather than a variable, split up the calculations into simple operations.

6 in binary is 110 which means \$1\times4 + 1\times2 + 0\times1 = 4 + 2\$. So how could you calculate \$6\times a\$ simply?

Hint: How do you multiply a number by 2 using only addition?

So how do you calculation \$6\times a\$ using only adders?

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    \$\begingroup\$ Hint: In binary you don't even need an adder to multiply by 2. \$\endgroup\$ – The Photon May 28 '16 at 2:49
  • \$\begingroup\$ Do you mean that multiplication is just a repetitive addition? \$\endgroup\$ – iMohaned May 28 '16 at 5:41
  • \$\begingroup\$ The problem is how to get the answer in BCD? \$\endgroup\$ – iMohaned May 28 '16 at 5:42

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