1.What would be the highest value of X in the given equation 4X1 5Y3 2Z7=1181 Explain this please.

Hello Gayatri

The equation is:

$4X1+5Y3+2Z7=1181$

Assume that $\mathbf{4}\mathbf{X}\mathbf{1},\mathbf{5}\mathbf{Y}$ , and $\mathbf{2}\mathbf{Z}$ 7 are three-digit numbers, where $X,Y$ , and $Z$ are digits (0-9).

Then:

• $4X1=401+10X$
• $5Y3=503+10Y$
• $2Z7=207+10Z$

Substituting these into the equation:

$(401+10X)+(503+10Y)+(207+10Z)=1181$

Simplify the constants:

$1111+10(X+Y+Z)=1181$

Subtract 1111 from both sides:

$10(X+Y+Z)=70$

Divide by 10:

$X+Y+Z=7$

Now we need the highest possible value of $X$ .

Since $X+Y+Z=7$ and $Y$ and $Z$ are digits (minimum value is 0 ),

To maximize $X$ , make $Y=0$ and $Z=0$ .

Then:

$X=7$

Verification

• $4X1=471$
• $5Y3=503$
• $2Z7=207$

Adding them:

$471+503+207=1181$

The equation is satisfied.