=TEXT(INT(1000000*5.84%/12),"0")&TEXT(MOD(1000000*5.84%/12,1),".000000000000")

IMHO, that works only by coincidence. In general, I agree with Fluff.

Consider the following example:

=1234000*5.78%/12

Excel displays only the first 15 significant digit result 5943.76666666667, rounded.

The calculation in VBA using type Decimal can return the 16 significant digit result 5943.766666666667, rounded. The result with the full precision of type Decimal is 5943.7666666666666666666666667.

But Tetra201's formula returns 5943.76666666666

6.

The reason is: In Excel, the exact decimal representation of the binary result of 1234000*5.78%/12 is 5943.76666666666,642413474619388580322265625.

So the exact decimal representation of the binary result of MOD(1234000*5.78%/12, 1) is 0.766666666666

424,13474619388580322265625.

(I use period for the decimal point and comma to demarcate the first 15 significant digits, which is all that Excel formats, rounded.)

Tetra201's formula might or might not agree with Yugaa2010's calculator. It depends on how the calculator performs arithmetic and represents the result. (The two might be different, as is the case in Intel-compatible computers.) 64-bit binary floating-point is only of several alternatives that are commonly used in modern calculators.

The Win7 Calculator displays 5943.766666666667 (16 significant digits) in Standard mode, which agrees with the VBA type Decimal result and differs from Tetra201's formula.

It displays 5943.7666666666666666666666666667 in Scientific mode (32 significant digits), which agrees with the VBA type Decimal result, but with 3 more digits.